Diagonalization and similarity

Proof of the finite real matrix product laws

Let $m,n,p,q$ be positive integers, let $A\in\mathbb R^{m\times n}$, $B\in\mathbb R^{n\times p}$, and $C\in\mathbb R^{p\times q}$. All the following arguments use the row-by-column definition and finite real sums. For any output position $(i,\ell)$,

$$\begin{aligned} ((AB)C)_{i\ell} &=\sum_{k=1}^{p}\left(\sum_{j=1}^{n}a_{ij}b_{jk}\right)c_{k\ell}\\ &=\sum_{j=1}^{n}a_{ij}\left(\sum_{k=1}^{p}b_{jk}c_{k\ell}\right) =(A(BC))_{i\ell}. \end{aligned}$$

The middle equality only reorders two finite sums and uses real distributivity. Both products are $m\times q$, so equality at every position proves associativity. No factor has changed its order.

For $D\in\mathbb R^{n\times p}$, the same definition gives

$$[A(B+D)]_{ik}=\sum_{j=1}^{n}a_{ij}(b_{jk}+d_{jk}) =(AB)_{ik}+(AD)_{ik}.$$

For $E\in\mathbb R^{m\times n}$, it gives

$$[(A+E)B]_{ik}=\sum_{j=1}^{n}(a_{ij}+e_{ij})b_{jk} =(AB)_{ik}+(EB)_{ik}.$$

Thus both distributive laws hold with the displayed matching sizes. For every real scalar $c$,

$$[(cA)B]_{ik}=\sum_j(ca_{ij})b_{jk} =c\sum_j a_{ij}b_{jk} =\sum_j a_{ij}(cb_{jk})=[A(cB)]_{ik},$$

proving $(cA)B=c(AB)=A(cB)$ by entrywise equality. In particular, for $x,y\in\mathbb R^n$ and real $s,t$,

$$[A(sx+ty)]_i=\sum_j a_{ij}(sx_j+ty_j) =s(Ax)_i+t(Ay)_i.$$

This proves the matrix-vector linearity used when comparing solutions of the same linear system. All identities concern compatible finite real matrices; they do not assert commutativity or cancellation of arbitrary matrix factors.

A vector space is a set with addition and scalar multiplication satisfying the usual abelian-group and distributive axioms over a field $F$. Columns are one model. Polynomials of degree at most $d$, denoted $P_d(F)$, and linear maps between spaces are others. The zero polynomial is included in $P_d(F)$ although its degree is not ordinarily defined.

A subspace contains zero and is closed under sums and scalar multiples. A list spans if every vector is a linear combination of its entries; it is independent if only the zero coefficients give zero. A basis is both. A finite spanning list can be reduced to a basis by deleting dependent entries; an independent list can be extended to a basis by adding vectors outside its span until the given finite spanning list is spanned.

Exchange lemma. An independent list of $r$ vectors in the span of $s$ vectors has $r\le s$. To see this, express the first independent vector in the spanning list, use a nonzero coefficient to replace one spanning vector by it, and repeat. At each stage independence ensures the next independent vector has a nonzero coefficient on an as-yet unreplaced spanning vector. There must be at least $r$ such slots. Comparing two bases in both directions proves that their lengths agree; this common length is $\dim V$.

Consequently, in dimension $n$, any $n$ independent vectors form a basis, as do any $n$ spanning vectors. Counting a list is useful only after its independence or spanning has been justified.

Finite basis existence and extension, without a counting assumption

Let $V$ be a real vector space supplied with a finite spanning list $t_1,\ldots,t_N$. First prove that every independent list $u_1,\ldots,u_k$ has length at most $N$. After $j$ insertions, maintain a spanning list consisting of $u_1,\ldots,u_j$ and $N-j$ surviving old vectors. Expand $u_{j+1}$ in this list. Some coefficient on a surviving old vector is nonzero; otherwise $u_{j+1}$ would be in the span of the earlier $u_i$, contradicting independence. Solve for that old vector and replace it by $u_{j+1}$. The new list spans the same space and keeps every earlier insertion. If $k>N$, after $N$ steps the first $N$ independent vectors already span $V$, contradicting independence of $u_{N+1}$. This proves the bound, including $N=0$.

Every subspace $W$ has a finite basis. Start with the empty independent list. If its current span is not $W$, choose one vector of $W$ outside that span. A relation with nonzero coefficient on the new vector would express it in the old span, so the enlarged list remains independent. The proved bound permits at most $N$ additions; therefore the process stops with an independent spanning list for $W$. If $W=\{0\}$, it stops immediately with the empty basis. Starting instead with any given independent list in $W$ proves that list extends to a basis of $W$. Only a bounded finite selection is used. Finally, if two bases have sizes $p,q$, use each as the finite spanning list in the bound to obtain $p\le q$ and $q\le p$. Thus basis size, called dimension, is well-defined. This supplies the existence, extension and dimension facts used in rank–nullity.

A map $T:V\to W$ is linear if $T(\alpha v+\beta u)=\alpha Tv+\beta Tu$. Its kernel and image are subspaces. Choose a basis $k_1,\ldots,k_r$ of $\ker T$ and extend to a basis with $v_1,\ldots,v_s$. Then $Tv_1,\ldots,Tv_s$ form a basis of $\operatorname{im}T$: spanning follows by expanding any input; a dependence among these images places the corresponding input combination in the kernel, and independence of the extended basis makes all coefficients zero. Thus

$$\dim V=\dim\ker T+\dim\operatorname{im}T.$$

This is rank–nullity. In equal finite dimensions, injectivity and surjectivity are equivalent, but the equivalence can fail in infinite-dimensional spaces.

One eigenvector is useful. A whole basis of eigenvectors is transformative. In this note the default scalar field is the real field $\mathbb{R}$: matrices, scalars, and vectors have real entries unless a complex extension is stated explicitly. Every argument is algebraic and also works over $\mathbb{C}$ after replacing $\mathbb{R}$ by $\mathbb{C}$.

If a square matrix has enough linearly independent eigenvectors, there is a coordinate system in which its action is diagonal. In that coordinate system, powers and inverses become entrywise calculations. The word “enough” is exact: for an $n\times n$ matrix, we need $n$ linearly independent eigenvectors, so those vectors form a basis of $\mathbb{R}^n$.

Similarity is a change of coordinates

The coordinate-conversion matrix is developed in Basis extension and change of basis.

Let $S$ be invertible. Write a vector in the original coordinates as $x=Sy$, where $y$ contains its coordinates in the basis formed by the columns of $S$. Set $B=S^{-1}AS$. Then

$$Ax=ASy=SBy.$$

Thus $A$ acting on the original vector $x$ is represented by $B$ acting on the new coordinate vector $y$. These two matrices describe one linear action in two bases. The matrix $S$ must be invertible before this interpretation makes sense: otherwise $y\mapsto Sy$ is not a coordinate system.

This viewpoint also explains why the equation is written with $S^{-1}$ on the left. Start with a vector $y$ in the new coordinates, convert it to $x=Sy$, apply the original matrix to obtain $ASy$, and convert the output back by $S^{-1}$. The resulting new-coordinate action is exactly $S^{-1}ASy$. A similarity calculation therefore preserves the map while changing the labels used for its inputs and outputs. It does not permit arbitrary replacement of one matrix by another: the conjugating matrix and its inverse record the coordinate conversion.

Definition

Similarity

Two $n\times n$ matrices $A$ and $B$ over the same field are similar if there exists an invertible $S$ such that

$$S^{-1}AS=B.$$

Definition

Diagonalization and diagonalizability

Let $A$ be an $n\times n$ matrix. If there is an invertible matrix $S$ and scalars $\lambda_1,\ldots,\lambda_n$ such that

$$S^{-1}AS=D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n),$$

then this equality is a diagonalization of $A$ and $A$ is diagonalizable.

Diagonalization is therefore a special similarity in which the target matrix has no off-diagonal entries. The diagonal entries are useful only when their order is tied to the columns of $S$.

The eigenvector dictionary

Theorem

Characterization of a displayed diagonalization

Let $A$ be $n\times n$ and let

$$S=[v_1\ v_2\ \cdots\ v_n]$$

be invertible. For scalars $\lambda_1,\ldots,\lambda_n$, set $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. The following statements are equivalent:

1. $Av_j=\lambda_jv_j$ for every $j=1,\ldots,n$;
2. $AS=SD$;
3. $S^{-1}AS=D$;
4. $A=SDS^{-1}$.

Proof

The four equivalent forms and their dependency

The column equation $Av_j=\lambda_jv_j$ for each $j$ is exactly the matrix identity $AS=SD$, because the $j$th columns of the two sides are those two vectors. Since $S$ is invertible, left multiplication by $S^{-1}$ gives $S^{-1}AS=D$; right multiplication by $S^{-1}$ gives $A=SDS^{-1}$. Conversely, each of those two equalities can be multiplied by $S$ on the appropriate side to recover $AS=SD$, and then the column equations. The invertibility of $S$ is used before either cancellation step, and each $v_j$ is nonzero automatically because it is a column of an invertible matrix.

The dictionary prevents a common ordering error. If the first column of $S$ is an eigenvector for $\lambda_1$ and the second is one for $\lambda_2$, then the diagonal is $\operatorname{diag}(\lambda_1,\lambda_2,\ldots)$. Reordering the columns requires reordering the diagonal entries at the same time.

There is a useful rectangular version of the first step. If $V=[v_1\ \cdots\ v_p]$ is an $n\times p$ matrix and $D=\operatorname{diag}(\mu_1,\ldots,\mu_p)$, then

$$AV=VD$$

is equivalent to $Av_j=\mu_jv_j$ for each listed column. This packages eigenvector equations even when $p<n$. It is not yet a diagonalization: a rectangular $V$ has no inverse, and dependent columns do not supply a full coordinate system. Diagonalization begins only when $p=n$ and the columns are independent. Keeping this distinction prevents the phrase “these vectors diagonalize $A$” from being applied to a partial eigenvector list.

Theorem

The exact criterion for diagonalizability

An $n\times n$ matrix over $\mathbb{R}$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors over $\mathbb{R}$. Equivalently, its eigenvectors can be chosen as a basis of $\mathbb{R}^n$.

For necessity, a diagonalization supplies the $n$ columns of an invertible $S$, and the dictionary says that each column is an eigenvector. For sufficiency, choose $n$ independent eigenvectors as columns of $S$. Independence makes $S$ invertible, and the dictionary supplies the diagonalization. Merely listing $n$ eigenvectors is not enough: repeated vectors or dependent choices do not make a coordinate basis.

In practice, a reliable construction has three separate checks. First solve $(A-\lambda I)v=0$ and record the field over which the nullspace is being taken. Next place the selected vectors in a matrix in an explicitly stated order and test independence, for example by checking that its determinant is nonzero. Finally multiply $AS$ and $SD$ or compute $S^{-1}AS$; one successful entry is not enough to certify a matrix identity. These checks separate the spectral question (which vectors exist) from the coordinate question (whether they form a basis) and from the arithmetic question (whether the displayed products were copied correctly).

First diagonalization examples

Worked example

Build $S$ and $D$ from a 2 by 2 matrix

Let

$$A=\begin{bmatrix}2&1\\1&2\end{bmatrix}.$$

Its characteristic polynomial is

$$\det(A-\lambda I)=(2-\lambda)^2-1 =\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3).$$

For $\lambda=3$, the equation $(A-3I)x=0$ gives $x_2=x_1$, so take $v_1=\begin{bmatrix}1\\1\end{bmatrix}$. For $\lambda=1$, the equation $(A-I)x=0$ gives $x_2=-x_1$, so take $v_2=\begin{bmatrix}1\\-1\end{bmatrix}$. Put them in that order:

$$S=\begin{bmatrix}1&1\\1&-1\end{bmatrix}, \qquad D=\begin{bmatrix}3&0\\0&1\end{bmatrix}.$$

The determinant of $S$ is $-2$, and

$$S^{-1}=\frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}=\frac12S.$$

The column equations give $AS=SD$, hence $S^{-1}AS=D$. For a coordinate check, take $x=\begin{bmatrix}3\\1\end{bmatrix}$. Since $x=Sy$ with $y=\begin{bmatrix}2\\1\end{bmatrix}$, we have

$$Dy=\begin{bmatrix}6\\1\end{bmatrix}, \qquad SDy=\begin{bmatrix}7\\5\end{bmatrix}=Ax.$$

The diagonal calculation and the original matrix action agree.

The first example also illustrates why a diagonalization is a basis statement, not merely a convenient factorization. The columns of $S$ give every vector a unique pair of new coordinates. In those coordinates the first direction is scaled by $3$ and the second by $1$; converting back with $S$ recombines those two independent effects. If the columns were dependent, some vectors would have no coordinates or would have more than one coordinate description, so the formula $S^{-1}AS$ would not represent a legitimate change of basis.

Worked example

A diagonalizable upper-triangular matrix

Let

$$A=\begin{bmatrix}1&1&1\\0&2&2\\0&0&3\end{bmatrix}.$$

The following eigenvectors are

$$u_1=\begin{bmatrix}1\\0\\0\end{bmatrix},\quad u_2=\begin{bmatrix}1\\1\\0\end{bmatrix},\quad u_3=\begin{bmatrix}3\\4\\2\end{bmatrix},$$

with eigenvalues $1,2,3$. Form

$$U=[u_1\ u_2\ u_3]=\begin{bmatrix}1&1&3\\0&1&4\\0&0&2\end{bmatrix}, \qquad D=\operatorname{diag}(1,2,3).$$

Its determinant is $2$, so it is invertible, with

$$U^{-1}=\begin{bmatrix}1&-1&1/2\\0&1&-2\\0&0&1/2\end{bmatrix}.$$

Direct multiplication gives

$$Au_1=\begin{bmatrix}1\\0\\0\end{bmatrix}=1u_1,\quad Au_2=\begin{bmatrix}2\\2\\0\end{bmatrix}=2u_2,\quad Au_3=\begin{bmatrix}9\\12\\6\end{bmatrix}=3u_3.$$

Thus $AU=UD$, and $U^{-1}AU=\operatorname{diag}(1,2,3)$. The upper-triangular shape of $A$ was not itself the proof of diagonalizability; the independent eigenvectors and the invertible matrix they form were.

Worked example

A repeated eigenvalue without a full eigenbasis

Consider

$$J=\begin{bmatrix}1&4\\0&1\end{bmatrix}.$$

Its only eigenvalue is $1$, since $\det(J-\lambda I)=(1-\lambda)^2$. Solving

$$(J-I)x=\begin{bmatrix}0&4\\0&0\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}=0$$

forces $x_2=0$, while $x_1$ is free. Thus the eigenspace is $\operatorname{span}\{\begin{bmatrix}1\\0\end{bmatrix}\}$, which has dimension one. A $2\times2$ matrix needs two independent eigenvectors, so $J$ is not diagonalizable. A repeated eigenvalue can still allow diagonalization, but its eigenspace must have enough independent directions.

Reordering and similarity invariants

If $Q$ is an $n\times n$ permutation matrix and $A=SDS^{-1}$, then $SQ$ is also invertible and

$$(SQ)^{-1}A(SQ)=Q^{-1}S^{-1}ASQ=Q^{-1}DQ.$$

The result is diagonal with the same entries reordered according to $Q$. This is why a diagonalization is not unique: eigenvectors may be placed in another simultaneous order. Rescaling an eigenvector column by a nonzero scalar also preserves its eigendirection, provided the corresponding column and diagonal ordering are kept aligned.

Theorem

Similarity preserves the characteristic polynomial

If $B=S^{-1}AS$ for invertible $S$, then

$$\det(B-tI)=\det\bigl(S^{-1}(A-tI)S\bigr)=\det(A-tI).$$

Therefore similar matrices have the same characteristic polynomial and the same eigenvalues, counted with algebraic multiplicity.

The calculation uses $S^{-1}(A-tI)S=S^{-1}AS-tS^{-1}IS$ and $S^{-1}IS=I$, followed by multiplicativity of the determinant. Similarity does not mean that the matrices have the same entries or the same eigenvectors; it means that corresponding eigenspaces are carried between coordinate systems by $S^{-1}$.

The characteristic-polynomial conclusion is one-way. It is safe to use equal characteristic polynomials as a quick obstruction: if the polynomials differ, the matrices cannot be similar. It is unsafe to reverse the implication, because the polynomial records aggregate eigenvalue information and can miss how eigenspaces and generalized directions are arranged. The nilpotent pair in the counterexample makes this limitation visible with the smallest nontrivial size.

Counterexample mode

Characteristic data alone does not prove similarity

The matrices

$$N=\begin{bmatrix}0&1\\0&0\end{bmatrix},\qquad 0_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$

both have characteristic polynomial $t^2$, but they are not similar. Every invertible matrix $S$ satisfies $S^{-1}0_2S=0_2$, whereas $N\ne0_2$. Equal characteristic polynomials are necessary for similarity, not sufficient.

Powers, inverse, and transpose

Theorem

Powers, inverse, and transpose from one diagonalization

Suppose $A=SDS^{-1}$, where $S$ is invertible and $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Then for every positive integer $m$,

$$A^m=SD^mS^{-1},\qquad D^m=\operatorname{diag}(\lambda_1^m,\ldots,\lambda_n^m).$$

If $A$ is invertible, every $\lambda_j$ is nonzero and $A^{-1}=SD^{-1}S^{-1}$. Also $A^T$ is diagonalizable with the same eigenvalues.

The positive-power statement starts at $m=1$; separately, $A^0=I_n$ and $D^0=I_n$. Negative powers require invertibility: for an integer $m<0$, the formula is $A^m=SD^mS^{-1}$ only when every diagonal entry is nonzero. A zero diagonal entry makes $D^{-1}$ and hence $A^{-1}$ undefined.

The boundary cases are worth keeping separate. A matrix with a zero diagonal entry in $D$ may still have well-defined nonnegative powers, including $A^0$, but it cannot have an inverse. If all diagonal entries are nonzero, then every integer power exists and the negative powers are obtained by replacing each $\lambda_j$ with its reciprocal. This is a statement about the same fixed diagonalization; one must not silently choose a different ordering for each power, because the conjugating factors would then no longer cancel in the displayed product.

Proof

Why the power and transpose formulas work

For $m+1$, multiply the induction hypothesis by $A$ and cancel the adjacent $S^{-1}S$:

$$A^{m+1}=(SD^mS^{-1})(SDS^{-1})=SD^{m+1}S^{-1}.$$

For the inverse, invert the three factors in reverse order: $(SDS^{-1})^{-1}=SD^{-1}S^{-1}$. For the transpose,

$$A^T=(S^{-1})^TD S^T.$$

Set $W=(S^{-1})^T$; then $W^{-1}=S^T$ and $W^{-1}A^TW=D^T=D$. Thus the columns of $W$ are eigenvectors of $A^T$. The original columns of $S$ need not be eigenvectors of $A^T$.

Worked example

Compute all positive powers of a symmetric matrix

Let

$$B=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix},$$

and take

$$S=\begin{bmatrix}1&1&1\\1&-1&0\\1&0&-1\end{bmatrix},\quad S^{-1}=\frac13\begin{bmatrix}1&1&1\\1&-2&1\\1&1&-2\end{bmatrix},\quad D=\operatorname{diag}(4,1,1).$$

The three columns of $S$ are eigenvectors, so $B=SDS^{-1}$. For every positive integer $m$,

$$B^m=\frac13\begin{bmatrix} 4^m+2&4^m-1&4^m-1\\ 4^m-1&4^m+2&4^m-1\\ 4^m-1&4^m-1&4^m+2 \end{bmatrix}.$$

At $m=1$ this gives diagonal entries $2$ and off-diagonal entries $1$, which recovers $B$. At $m=2$, the diagonal entries are $6$ and the off-diagonal entries are $5$, matching direct multiplication. The symbolic power is useful because no new eigenvector calculation is needed for another positive $m$.

A complex extension and common eigenbases

Over $\mathbb{R}$, the matrix

$$R=\begin{bmatrix}1&-1\\1&1\end{bmatrix}$$

has characteristic polynomial $(1-\lambda)^2+1$, so it has no real eigenvalues and is not diagonalizable over $\mathbb{R}$. Over $\mathbb{C}$, use

$$U=\begin{bmatrix}i&-i\\1&1\end{bmatrix},\quad U^{-1}=\frac12\begin{bmatrix}-i&1\\i&1\end{bmatrix},\quad D=\operatorname{diag}(1+i,1-i).$$

Then $U^{-1}RU=D$; the complex diagonalization is valid because $\det(U)=2i\ne0$. This is a field extension, not a contradiction of the real criterion.

Theorem

Sums and products with a common eigenbasis

Suppose $A=SDS^{-1}$ and $B=SES^{-1}$ for the same invertible $S$, where $D$ and $E$ are diagonal. For any scalars $\alpha,\beta$,

$$\alpha A+\beta B=S(\alpha D+\beta E)S^{-1}, \qquad AB=SDE S^{-1}.$$

Both right-hand middle matrices are diagonal, so these matrices are simultaneously diagonalizable by $S$.

Indeed, distribute and use $S^{-1}S=I$: $\alpha SDS^{-1}+\beta SES^{-1}=S(\alpha D+\beta E)S^{-1}$ and $(SDS^{-1})(SES^{-1})=SDE S^{-1}$. The same eigenbasis hypothesis is essential; separate diagonalizability of $A$ and $B$ does not by itself give a common $S$.

What repeated eigenvalues do and do not tell us

The characteristic polynomial counts algebraic multiplicity, while the dimension of an eigenspace counts available independent eigenvectors. These numbers are related but answer different questions. If a repeated eigenvalue has an eigenspace whose dimension is too small, no rearrangement of columns can create the missing directions. The matrix $J$ demonstrates this obstruction. If the eigenspace has enough dimension, repeated diagonal entries cause no problem; a diagonal matrix itself may have the same eigenvalue many times and still have a complete eigenbasis.

Similarity preserves the characteristic polynomial because it preserves the linear map under a coordinate conversion. It also carries an eigenspace for $A$ to the corresponding eigenspace for $B$: if $Av=\lambda v$ and $B=S^{-1}AS$, then $B(S^{-1}v)=\lambda(S^{-1}v)$. The vector remains nonzero because $S^{-1}$ is invertible. This explains why eigenvalues agree while the literal eigenvectors usually differ. It also clarifies the limitation of the polynomial invariant: the polynomial records eigenvalue data, not the full arrangement of eigenspaces.

Check the final certificate

To decide diagonalizability over the stated field, find every eigenvalue and add the dimensions of its eigenspace. The total is $n$ exactly when an $n\times n$ matrix has an eigenbasis. Assemble an ordered basis as the columns of $S$, use the same eigenvalue order in $D$, and verify $AS=SD$ together with the invertibility of $S$. These two checks establish the claimed diagonalization without requiring a second computation of $S^{-1}$.

Quick checks and exercises

Checkpoint

Suppose $S$ is invertible and $S^{-1}AS=D$ is diagonal. How are the columns of $S$ related to the eigenvectors of $A$ and the diagonal entries of $D$?

Use the equation $AS=SD$.

Solution · Answer

The columns must be linearly independent eigenvectors of $A$, forming an eigenbasis; the $j$th column corresponds to the $j$th diagonal entry of $D$.

Checkpoint

Can the maximum number of linearly independent eigenvectors of a diagonalizable 3×3 matrix be two?

Use the exact criterion.

Solution · Answer

No. If it is diagonalizable, its maximum number of linearly independent eigenvectors is three, so it cannot be two.

Checkpoint

If $A$ is similar to a diagonal matrix $D$, do $A$ and $D$ have the same eigenvalues?

Use the characteristic-polynomial invariant.

Solution · Answer

Yes. Similar matrices have the same characteristic polynomial and therefore the same eigenvalues, with algebraic multiplicity.

Checkpoint

Suppose $A$ has eigenvectors $v_1,v_2,v_3$ with eigenvalues $2,5,-1$, and these vectors are linearly independent. What diagonal matrix appears in a diagonalization of the $3×3$ matrix $A$?

Match the diagonal order to the column order.

Solution · Guided solution

If $S=[v_1\ v_2\ v_3]$, then

$$S^{-1}AS=\operatorname{diag}(2,5,-1).$$

The entries must match the eigenvalues attached to the selected columns.

Checkpoint

Why is $J=\begin{bmatrix}1&4\\0&1\end{bmatrix}$ not diagonalizable?

Find the dimension of its eigenspace.

Solution · Guided solution

Its only eigenvalue is $1$, and $(J-I)x=0$ forces $x_2=0$. The eigenspace is one-dimensional, but a $2\times2$ matrix needs two independent eigenvectors.

Checkpoint

If $A=SDS^{-1}$ with $D=\operatorname{diag}(2,-1)$, what is $A^3$?

Use the positive-power formula.

Solution · Guided solution

$$A^3=SD^3S^{-1}=S\operatorname{diag}(8,-1)S^{-1}.$$

Related notes

Read 8.1 Eigenvalues, eigenvectors, and eigenspaces first if the homogeneous-system viewpoint is not yet solid. Continue with 8.3 Characteristic polynomials and diagonalization tests for polynomial tools that test whether enough eigenvectors exist. The basis language here also depends on 6.5 Basis and dimension.