Exercises — Diagonalization — conditions, procedure

A diagonal matrix is already diagonal, so trivially $A=PD{P}^{-1}$ with $P=I$ (the identity) and $D$ itself.

• WHAT we did: recognized the form. WHY it works: $I$ is invertible and $ID{I}^{-1}=D$.
• The standard basis vectors $\left(\begin{array}{c}1\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ 1\end{array}\right)$ are the eigenvectors, with eigenvalues $5$ and $2$.

Answer: yes; $P=I$, $D=\left(\begin{array}{cc}5& 0\\ 0& 2\end{array}\right)$.

$A$ is upper triangular, so its eigenvalues sit on the diagonal (the characteristic polynomial is $(7-\lambda )(1-\lambda )$, see Characteristic Polynomial). The eigenvalues are $7$ and $1$ — two distinct values.

• WHY that settles it: distinct eigenvalues force linearly independent eigenvectors, and $2$ independent eigenvectors in ${\mathbb{R}}^2$ span it.

Answer: yes, guaranteed diagonalizable (distinct eigenvalues $7,1$).

False. All three eigenvalues equal $4$ (algebraic multiplicity $a(4)=3$). Diagonalizability now depends entirely on the geometric multiplicity $g(4)=\dim \ker (A-4I)$.

• If $g(4)=3$ (i.e. $A=4I$), diagonalizable.
• If $g(4)<3$ (defective), not diagonalizable.

The polynomial alone cannot tell us. Answer: False.

Step 1 — eigenvalues. $\det (A-\lambda I)=(1-\lambda {)}^2-4={\lambda }^2-2\lambda -3=(\lambda -3)(\lambda +1)$. So ${\lambda }_1=3,{\lambda }_2=-1$. Distinct → diagonalizable. Step 2 — eigenvectors. $\lambda =3$: $A-3I=\left(\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right)\Rightarrow x_1=x_2\Rightarrow {\mathbf{p}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$. $\lambda =-1$: $A+I=\left(\begin{array}{cc}2& 2\\ 2& 2\end{array}\right)\Rightarrow x_1=-x_2\Rightarrow {\mathbf{p}}_2=\left(\begin{array}{c}1\\ -1\end{array}\right)$. Step 3 — assemble (order matched): $P=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right),D=\left(\begin{array}{cc}3& 0\\ 0& -1\end{array}\right).$ $\det P=-2\ne 0$ so $P$ is invertible. ✓ (This $A$ is symmetric, so eigenvectors come out orthogonal — see Spectral Theorem.)

Eigenvalues: $(5-\lambda {)}^2=0\Rightarrow \lambda =5$, so $a(5)=2$. Eigenvectors: $A-5I=\left(\begin{array}{cc}0& 4\\ 0& 0\end{array}\right)$. Solving gives $x_2=0$, $x_1$ free → one direction $\left(\begin{array}{c}1\\ 0\end{array}\right)$, so $g(5)=1$. Since $g(5)=1<a(5)=2$, not diagonalizable (defective — it is a scaled shear).

WHY diagonalize first? $A^4=P{D}^4{P}^{-1}$ and $D^4$ is just each diagonal entry to the 4th power (this is the whole point of Matrix Powers and Exponentials). $D^4=\left(\begin{array}{cc}{3}^4& 0\\ 0& (-1{)}^4\end{array}\right)=\left(\begin{array}{cc}81& 0\\ 0& 1\end{array}\right)$. $P^{-1}=\frac{1}{-2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)$. $A^4=P{D}^4{P}^{-1}=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}81& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{cc}41& 40\\ 40& 41\end{array}\right).$

Eigenvalue: $(3-\lambda {)}^2=0\Rightarrow \lambda =3$, $a(3)=2$ for every $t$. Eigenvectors: $A-3I=\left(\begin{array}{cc}0& t\\ 0& 0\end{array}\right)$.

• If $t\ne 0$: solving forces $x_2=0$, $x_1$ free → $g(3)=1<2$ → not diagonalizable.
• If $t=0$: $A-3I=0$, whole plane is the eigenspace → $g(3)=2$ → diagonalizable (it's $3I$).

Answer: diagonalizable only when $t=0$.

Characteristic polynomial: $\det (R-\lambda I)={\lambda }^2+1=0\Rightarrow \lambda =\pm i$.

• Over $\mathbb{R}$: no real eigenvalue exists, so there is no real eigenvector — geometrically a rotation turns every real direction, so no real line is merely stretched. Not diagonalizable over $\mathbb{R}$. The figure below makes this concrete: it draws several black unit vectors and, in red, one vector $v$ together with its image $Rv$ turned a full $90^{\circ }$ away — since no black arrow lands back on its own line, there is no direction that is only stretched, i.e. no real eigenvector.
• Over $\mathbb{C}$: two distinct eigenvalues $i,-i$ → diagonalizable, with $D=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right).$ Eigenvectors: for $\lambda =i$, $(R-iI)\mathbf{x}=0$. The first row gives $-i{x}_1-x_2=0\Rightarrow x_2=-i{x}_1$, so take ${\mathbf{p}}_1=\left(\begin{array}{c}1\\ -i\end{array}\right)$ (check: $R{\mathbf{p}}_1=\left(\begin{array}{c}i\\ 1\end{array}\right)=i\left(\begin{array}{c}1\\ -i\end{array}\right)$ ✓). Likewise for $\lambda =-i$, ${\mathbf{p}}_2=\left(\begin{array}{c}1\\ i\end{array}\right)$.

Check $g=a$ for each eigenvalue:

• $\lambda =2$: $a(2)=2$ and $g(2)=2$ → matches.
• $\lambda =5$: $a(5)=1$, and always $1\le g\le a$, so $g(5)=1$ → matches.

Total independent eigenvectors $=2+1=3=n$. Diagonalizable. $P$ gets 2 columns from the $\lambda =2$ eigenspace and 1 column from the $\lambda =5$ eigenspace.

Lower triangular → eigenvalues on the diagonal: $\lambda =2,2,3$. So $a(2)=2,a(3)=1$. Eigenvectors for $\lambda =2$: $A-2I=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 1\end{array}\right)$. Equations: $x_1=0$ and $x_2+x_3=0$. Free variable: one → $\mathbf{p}=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$. So $g(2)=1$. Since $g(2)=1<a(2)=2$, not diagonalizable (defective — a Jordan block sits at $\lambda =2$; see Jordan Normal Form).

Determinant. $\det A=\det (PD{P}^{-1})=\det P\cdot \det D\cdot \det (P^{-1})$. Since $\det (P^{-1})=1/\det P$, the $\det P$ factors cancel, leaving $\det A=\det D={\prod }_i{\lambda }_i$ (determinant of a diagonal matrix is the product of its diagonal). Trace. Trace is cyclic: $\mathrm{tr}(XY)=\mathrm{tr}(YX)$. So $\mathrm{tr}A=\mathrm{tr}(PD{P}^{-1})=\mathrm{tr}(D{P}^{-1}P)=\mathrm{tr}(D)={\sum }_i{\lambda }_i.$ Both facts follow purely from $A$ being similar to $D$ (see Similar Matrices) — indeed they hold for any matrix via its eigenvalues.

WHY diagonalize: $e^A=P{e}^D{P}^{-1}$, and $e^D$ of a diagonal matrix is just $e$ of each diagonal entry (this defines the exponential termwise; see Matrix Powers and Exponentials). $e^D=\left(\begin{array}{cc}{e}^3& 0\\ 0& e^{-1}\end{array}\right)$, and with $P=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$, $P^{-1}=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)$: $e^A=\left(\begin{array}{cc}\frac{e^3+e^{-1}}{2}& \frac{e^3-e^{-1}}{2}\\ \frac{e^3-e^{-1}}{2}& \frac{e^3+e^{-1}}{2}\end{array}\right).$

Strategy: take a Jordan block $J=\left(\begin{array}{cc}4& 1\\ 0& 4\end{array}\right)$ (eigenvalue $4$, defective, eigenvector $e_1=\left(\begin{array}{c}1\\ 0\end{array}\right)$) and rotate its eigen-direction to $\left(\begin{array}{c}1\\ 1\end{array}\right)$ via a change of basis $S=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ (its first column is the target eigenvector). Set $A=SJ{S}^{-1}$. $S^{-1}=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)$. $A=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)\left(\begin{array}{cc}4& 1\\ 0& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}3& 1\\ -1& 5\end{array}\right).$ Check eigenvalue: $\det (A-\lambda I)=(3-\lambda )(5-\lambda )+1={\lambda }^2-8\lambda +16=(\lambda -4{)}^2$ → $a(4)=2$. ✓ Check eigenvector: $A\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}4\\ 4\end{array}\right)=4\left(\begin{array}{c}1\\ 1\end{array}\right)$. ✓ Geometric multiplicity: $A-4I=\left(\begin{array}{cc}-1& 1\\ -1& 1\end{array}\right)$ has rank $1$, so $g(4)=2-1=1<2$. Not diagonalizable.

Answer: $A=\left(\begin{array}{cc}3& 1\\ -1& 5\end{array}\right)$.

Eigenvalues. If $A\mathbf{v}=\lambda \mathbf{v}$ with $\mathbf{v}\ne 0$, then $A^2\mathbf{v}={\lambda }^2\mathbf{v}$. But $A^2=A$ gives $A^2\mathbf{v}=A\mathbf{v}=\lambda \mathbf{v}$. So ${\lambda }^2=\lambda \Rightarrow \lambda (\lambda -1)=0\Rightarrow \lambda \in \{0,1\}$. Diagonalizable. For any vector $\mathbf{x}$ write the identity $\mathbf{x}=\underset{\in \text{range}}{\underbrace{A\mathbf{x}}}+\underset{\in \ker A}{\underbrace{(\mathbf{x}-A\mathbf{x})}}.$ The first piece is a $\lambda =1$ eigenvector: $A(A\mathbf{x})=A^2\mathbf{x}=A\mathbf{x}$. The second is a $\lambda =0$ eigenvector: $A(\mathbf{x}-A\mathbf{x})=A\mathbf{x}-A^2\mathbf{x}=0$. So every vector splits into eigenvectors → the eigenvectors span the whole space → $A$ is diagonalizable with $D=\mathrm{diag}(1,\dots ,1,0,\dots ,0)$. This is a clean special case of the Spectral Theorem intuition: projections are perfectly diagonalizable.

General reasoning. $A^k=P{D}^k{P}^{-1}$, and $D^k=\mathrm{diag}({\lambda }_1^k,\dots ,{\lambda }_n^k)$. Each entry's fate is governed by $|{\lambda }_i|$:

• $|{\lambda }_i|<1$: ${\lambda }_i^k\to 0$.
• ${\lambda }_i=1$: ${\lambda }_i^k=1$ stays put.
• $|{\lambda }_i|>1$ or ${\lambda }_i=-1$: ${\lambda }_i^k$ diverges/oscillates → no convergence.

So $A^k$ converges (to a possibly nonzero limit) exactly when each eigenvalue is either inside the unit circle or equal to $1$. Apply to $A=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$. From L2.1 its eigenvalues are ${\lambda }_1=3$ and ${\lambda }_2=-1$.

• ${\lambda }_1=3$ has $|3|=3>1$, so the diagonal entry $3^k\to \infty$ (it explodes).
• ${\lambda }_2=-1$ has $(-1{)}^k$ oscillating $1,-1,1,-1,\dots$ (it never settles).

Either one alone already breaks convergence. Since $A^k=P{D}^k{P}^{-1}$ carries these divergent diagonal entries, $A^k$ does NOT converge — it blows up (dominated by the $3^k$ term). Conclusion: no limit exists.