Corrected exercises: eigenvalues and diagonalization (201-SN4-RE)

Diagonal matrix: $\det(A-\lambda I)=(-3-\lambda)(7-\lambda)$, so the eigenvalues are the diagonal entries, $7$ and $-3$.

Triangular matrix: $\det(A-\lambda I)=(6-\lambda)(-1-\lambda)$, because the $-4$ does not enter the product of the diagonal. Eigenvalues $6$ and $-1$. Pitfall: bringing the $-4$ into the calculation.

$\operatorname{tr}=5+1=6$ and $\det=5(1)-(-2)(3)=11$, so $\lambda^{2}-6\lambda+11$: $b=-6$ and $c=11$. Pitfall: $b$ carries the MINUS sign of the trace, and $\det=5+6$, not $5-6$. The discriminant $36-44<0$ announces complex eigenvalues.

$\operatorname{tr}=4$ and $\det=3-8=-5$, so $\lambda^{2}-4\lambda-5=(\lambda-5)(\lambda+1)$: eigenvalues $5$ and $-1$. Check: $5+(-1)=4$ and $5\times(-1)=-5$.

$\operatorname{tr}=-4$ and $\det=-12-9=-21$, so $\lambda^{2}+4\lambda-21=(\lambda+7)(\lambda-3)$: eigenvalues $3$ and $-7$, real, as for every symmetric matrix.

$\operatorname{tr}=6$ and $\det=8+1=9$, so $\lambda^{2}-6\lambda+9=(\lambda-3)^{2}$: double eigenvalue $3$. Then $A-3I=\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}$, of rank $1$, so $\dim E_{3}=2-1=1$. Pitfall: believing that the dimension equals the multiplicity $2$; here it is $1$, and the matrix is not diagonalizable.

$\operatorname{tr}=4$ and $\det=3+10=13$, so $\lambda^{2}-4\lambda+13=0$, with discriminant $16-52=-36$. Hence $\lambda=\frac{4\pm 6i}{2}=2\pm 3i$: real part $2$, imaginary part $3$. Pitfall: forgetting to divide $6i$ by $2$.

Upper triangular matrix: the eigenvalues are $2$, $-4$ and $5$. The largest is $5$, the smallest $-4$.

Block matrix: $\det(A-\lambda I)=(4-\lambda)\left[(2-\lambda)(-1-\lambda)-4\right]=(4-\lambda)(\lambda^{2}-\lambda-6)=(4-\lambda)(\lambda-3)(\lambda+2)$. Eigenvalues $4$, $3$ and $-2$: the largest is $4$, the smallest $-2$. Pitfall: forgetting the $1\times 1$ block, which contributes $4$.

$A\vec{v}=\begin{pmatrix} 2-1+2 \\ 1-2-2 \\ 2+0+4 \end{pmatrix}=\begin{pmatrix} 3 \\ -3 \\ 6 \end{pmatrix}=3\vec{v}$: the eigenvalue is $3$. No characteristic polynomial is needed when the eigenvector is given.

Compute $A\vec{v}=\begin{pmatrix}3&1\\1&3\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}3+1\\1+3\end{pmatrix}=\begin{pmatrix}4\\4\end{pmatrix}=4\begin{pmatrix}1\\1\end{pmatrix}=4\vec{v}$. The relation $A\vec{v}=\lambda\vec{v}$ holds with $\lambda=4$: $\vec{v}$ is indeed an eigenvector, with eigenvalue 4. Geometrically, the matrix multiplies by 4 every vector pointing along the diagonal $y=x$, without deflecting it.

$A\vec{w}=\begin{pmatrix}3&1\\1&3\end{pmatrix}\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}3-1\\1-3\end{pmatrix}=\begin{pmatrix}2\\-2\end{pmatrix}=2\vec{w}$. So $\vec{w}$ is an eigenvector with eigenvalue 2. About the directions: $\vec{v}\cdot\vec{w}=(1)(1)+(1)(-1)=0$, the two eigenvectors are ORTHOGONAL. This is no accident: $A$ is symmetric ($a_{12}=a_{21}=1$), and the eigenvectors of a symmetric matrix associated with distinct eigenvalues are always orthogonal. Exercise 8 comes back to this.

$A\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}$. For this to be an eigenvector we would need $\begin{pmatrix}3\\1\end{pmatrix}=\lambda\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}\lambda\\0\end{pmatrix}$, which requires both $\lambda=3$ and $1=0$: impossible. So $\begin{pmatrix}1\\0\end{pmatrix}$ is NOT an eigenvector. Geometrically, $A$ turns this vector as well as lengthening it, since the image $\begin{pmatrix}3\\1\end{pmatrix}$ is not parallel to the original vector. The practical test is therefore simple: compute $A\vec{u}$ and check whether the result is PROPORTIONAL to $\vec{u}$.

Without the condition $\vec{v}\neq\vec{0}$, the definition would lose all content: $A\vec{0}=\vec{0}=\lambda\vec{0}$ holds for EVERY scalar $\lambda$, so any number would be an eigenvalue of any matrix. The notion would no longer distinguish anything. This is also why eigenspaces are defined as NULL SPACES: $E_{\lambda}=\ker(A-\lambda I)$ necessarily contains $\vec{0}$ (every subspace does), and the condition “$\lambda$ is an eigenvalue” translates exactly into “this null space contains SOMETHING OTHER than $\vec{0}$”, that is $\dim E_{\lambda}\geq 1$, or again $\det(A-\lambda I)=0$. Here you see why the determinant appears: it vanishes precisely when the homogeneous system has non-trivial solutions.

By definition, $\lambda$ is an eigenvalue if there is a $\vec{v}\neq\vec{0}$ with $A\vec{v}=\lambda\vec{v}$, which reads $A\vec{v}-\lambda\vec{v}=\vec{0}$, that is $(A-\lambda I)\vec{v}=\vec{0}$ (inserting $I$ is needed for the subtraction to make sense as a matrix operation). The question therefore becomes: does the homogeneous system $(A-\lambda I)\vec{v}=\vec{0}$ have a NON-TRIVIAL solution? A square homogeneous system has non-trivial solutions if and only if its matrix is not invertible, that is, if and only if its determinant is zero. Hence the equivalence $\lambda$ eigenvalue $\iff\det(A-\lambda I)=0$ ✓. The whole chapter rests on this translation, and it reuses directly the invertibility criterion from the determinants chapter.

$A-\lambda I=\begin{pmatrix}4-\lambda&2\\1&3-\lambda\end{pmatrix}$, with determinant $(4-\lambda)(3-\lambda)-2=12-4\lambda-3\lambda+\lambda^{2}-2=\lambda^{2}-7\lambda+10$. Setting it to zero: $\lambda^{2}-7\lambda+10=0$, that is $(\lambda-2)(\lambda-5)=0$. The eigenvalues are therefore $\lambda_{1}=2$ and $\lambda_{2}=5$.

Trace: $\operatorname{tr}(A)=4+3=7$, and $\lambda_{1}+\lambda_{2}=2+5=7$ ✓. Determinant: $\det(A)=4\times 3-2\times 1=10$, and $\lambda_{1}\lambda_{2}=2\times 5=10$ ✓. These two relations are simply the coefficients of the characteristic polynomial read off, since it always reads $\lambda^{2}-\operatorname{tr}(A)\lambda+\det(A)$ for a $2\times 2$ matrix. They give a five-second check, and even let you guess the eigenvalues when they are integers: here, looking for two numbers with sum 7 and product 10 gives 2 and 5 at once. This is the check to make every time on the exam.

The matrix $B$ is LOWER TRIANGULAR (all the entries above the diagonal are zero). The determinant of a triangular matrix is the product of its diagonal entries, so $\det(B-\lambda I)=(2-\lambda)(3-\lambda)(-1-\lambda)$, which vanishes for $\lambda=2$, $\lambda=3$ and $\lambda=-1$. The eigenvalues of a triangular matrix are therefore exactly its DIAGONAL entries, with no calculation at all. The same holds for upper triangular matrices and, as a limiting case, for diagonal matrices. This fits the goal of the chapter: diagonalizing a matrix means bringing it to a form where its eigenvalues are read directly off the diagonal.

$A-2I=\begin{pmatrix}2&2\\1&1\end{pmatrix}$. The system $(A-2I)\vec{v}=\vec{0}$ reads $2x+2y=0$ and $x+y=0$: the two equations are proportional and reduce to $x=-y$. Setting $y=t$: $\vec{v}=(-t,t)=t(-1,1)$. So $E_{2}=\operatorname{span}\left\{\begin{pmatrix}-1\\1\end{pmatrix}\right\}$, a line, and $\dim E_{2}=1$. A useful consistency check: the system MUST be degenerate (two proportional equations), otherwise the only solution would be $\vec{0}$ and $\lambda=2$ would not be an eigenvalue. If the row reduction does not degenerate, there is a mistake in the characteristic polynomial.

$A-5I=\begin{pmatrix}-1&2\\1&-2\end{pmatrix}$. The system gives $-x+2y=0$, that is $x=2y$. Setting $y=t$: $\vec{v}=(2t,t)=t(2,1)$. So $E_{5}=\operatorname{span}\left\{\begin{pmatrix}2\\1\end{pmatrix}\right\}$ and $\dim E_{5}=1$. Here again the two rows are proportional ✓.

Check for $\lambda=2$: $A\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}-4+2\\-1+3\end{pmatrix}=\begin{pmatrix}-2\\2\end{pmatrix}=2\begin{pmatrix}-1\\1\end{pmatrix}$ ✓. For $\lambda=5$: $A\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}8+2\\2+3\end{pmatrix}=\begin{pmatrix}10\\5\end{pmatrix}=5\begin{pmatrix}2\\1\end{pmatrix}$ ✓. Do they form a basis of $\mathbb{R}^{2}$? There are two of them in a space of dimension 2, so it is enough to check independence: the determinant $\begin{vmatrix}-1&2\\1&1\end{vmatrix}=-1-2=-3\neq 0$, they are independent and do form a BASIS. This is the fact that will make $A$ diagonalizable: there is a basis of $\mathbb{R}^{2}$ made entirely of eigenvectors.

An eigenvector is never unique because every non-zero multiple of an eigenvector is still an eigenvector for the same eigenvalue: if $A\vec{v}=\lambda\vec{v}$, then $A(c\vec{v})=cA\vec{v}=c\lambda\vec{v}=\lambda(c\vec{v})$ for every $c\neq 0$ ✓. Geometrically, it is the DIRECTION that is an eigendirection, not a particular vector in it. More generally, the set of eigenvectors associated with $\lambda$, together with $\vec{0}$, is exactly the subspace $E_{\lambda}=\ker(A-\lambda I)$: it is therefore closed under addition and scalar multiplication. The practical consequence on the exam: two students can give $\begin{pmatrix}-1\\1\end{pmatrix}$ and $\begin{pmatrix}3\\-3\end{pmatrix}$ as the eigenvector and both be right. Only the direction counts, which is why the expected answer is a BASIS of the eigenspace rather than one imposed vector.

Put the eigenvectors as the COLUMNS of $P$ and the corresponding eigenvalues on the diagonal of $D$, in the same order. With $\vec{v}_{2}=\begin{pmatrix}-1\\1\end{pmatrix}$ for $\lambda=2$ and $\vec{v}_{5}=\begin{pmatrix}2\\1\end{pmatrix}$ for $\lambda=5$: $P=\begin{pmatrix}-1&2\\1&1\end{pmatrix}$ and $D=\begin{pmatrix}2&0\\0&5\end{pmatrix}$.

$\det P=(-1)(1)-(2)(1)=-3$, so $P^{-1}=\frac{1}{-3}\begin{pmatrix}1&-2\\-1&-1\end{pmatrix}=\begin{pmatrix}-\frac{1}{3}&\frac{2}{3}\\\frac{1}{3}&\frac{1}{3}\end{pmatrix}$. Check: $PD=\begin{pmatrix}-1&2\\1&1\end{pmatrix}\begin{pmatrix}2&0\\0&5\end{pmatrix}=\begin{pmatrix}-2&10\\2&5\end{pmatrix}$. Then $PDP^{-1}=\begin{pmatrix}-2&10\\2&5\end{pmatrix}\begin{pmatrix}-\frac{1}{3}&\frac{2}{3}\\\frac{1}{3}&\frac{1}{3}\end{pmatrix}=\begin{pmatrix}\frac{2}{3}+\frac{10}{3}&-\frac{4}{3}+\frac{10}{3}\\-\frac{2}{3}+\frac{5}{3}&\frac{4}{3}+\frac{5}{3}\end{pmatrix}=\begin{pmatrix}4&2\\1&3\end{pmatrix}=A$ ✓.

If you swap the columns of $P$ to get $P'=\begin{pmatrix}2&-1\\1&1\end{pmatrix}$ while keeping $D=\begin{pmatrix}2&0\\0&5\end{pmatrix}$, the equality fails: you would be claiming that $\begin{pmatrix}2\\1\end{pmatrix}$ goes with the eigenvalue 2, when it goes with 5. The calculation confirms it, $P'DP'^{-1}=\begin{pmatrix}3&-2\\-1&4\end{pmatrix}\neq A$. The rule is therefore strict: the $i$-th column of $P$ must be an eigenvector associated with the $i$-th diagonal entry of $D$. On the other hand, swapping the columns of $P$ AND the entries of $D$ together is perfectly legal and gives another diagonalization, just as valid: diagonalization is not unique, only the PAIRING counts.

Read from right to left on a vector $\vec{x}$, the equality $A\vec{x}=PDP^{-1}\vec{x}$ breaks into three steps. First $P^{-1}$ CHANGES THE BASIS: it expresses $\vec{x}$ in the basis of eigenvectors, that is, it gives the coordinates of $\vec{x}$ along the two preferred directions. Next $D$ acts in that basis, and its action is as simple as possible: it multiplies the first coordinate by 2 and the second by 5, with no mixing at all, since $D$ is diagonal. Finally $P$ GOES BACK to the standard basis. The complicated transformation $A$ is therefore a plain double stretch, provided you look at it in the right frame. That is the whole point of the chapter: $A$ mixes the coordinates only because it is observed in a badly chosen basis, and diagonalizing amounts to choosing the frame in which the transformation becomes transparent.

The characteristic polynomial is $\det(A-\lambda I)=\begin{vmatrix} -\lambda & -1 \\ 1 & -\lambda \end{vmatrix}=\lambda^{2}+1$. Setting it to zero: $\lambda^{2}=-1$, so $\lambda=i$ and $\lambda=-i$. Both eigenvalues are purely imaginary and conjugate to each other. Note that $\operatorname{tr}(A)=0=i+(-i)$ and $\det(A)=1=i\cdot(-i)$.

A real eigenvector would be a non-zero real vector whose direction is preserved by $A$. But $A$ is a rotation through $90^{\circ}$: it turns EVERY vector of the plane by a quarter turn, so no non-zero real vector keeps its direction. There can therefore be no real eigenvector, which is consistent with the absence of a real eigenvalue. Eigendirections only appear once you move to the complex plane.

$\det(B-\lambda I)=\begin{vmatrix} 2-\lambda & -1 \\ 1 & 2-\lambda \end{vmatrix}=(2-\lambda)^{2}+1=\lambda^{2}-4\lambda+5$. The discriminant is $16-20=-4<0$, so the roots are complex: $\lambda=\dfrac{4\pm\sqrt{-4}}{2}=\dfrac{4\pm 2i}{2}=2\pm i$. The eigenvalues are $2+i$ and $2-i$.

Sum: $(2+i)+(2-i)=4=\operatorname{tr}(B)$ ✓. Product: $(2+i)(2-i)=4-i^{2}=4+1=5=\det(B)$ ✓. The two relations, proved for real eigenvalues, remain true for complex eigenvalues, because they are only the coefficients of the characteristic polynomial $\lambda^{2}-\operatorname{tr}(B)\lambda+\det(B)$ read off. Note that the common real part, $2$, is half the trace: this is general for a conjugate pair, and it gives a quick check. The matrix $B$ acts as a rotation combined with a scaling, which the link with complex numbers of modulus $\sqrt{5}$ will confirm.

ALGEBRAIC multiplicity of $\lambda$: its multiplicity as a root of the characteristic polynomial. GEOMETRIC multiplicity of $\lambda$: the dimension of the eigenspace $E_{\lambda}=\ker(A-\lambda I)$, in other words the number of independent eigenvectors associated with $\lambda$. Always, $1\leq$ geometric $\leq$ algebraic. CRITERION: $A$ of size $n$ is diagonalizable if and only if, for EVERY eigenvalue, the geometric multiplicity equals the algebraic multiplicity; equivalently, if and only if the dimensions of the eigenspaces add up to $n$, that is, if there is a basis of the space made of eigenvectors.

$A$ is triangular, so its eigenvalues are its diagonal entries: $\lambda=3$, with ALGEBRAIC multiplicity 2 (the characteristic polynomial is $(3-\lambda)^{2}$). Geometric multiplicity: $A-3I=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, and the system $(A-3I)\vec{v}=\vec{0}$ gives the single equation $y=0$, with $x$ free. So $E_{3}=\operatorname{span}\left\{\begin{pmatrix}1\\0\end{pmatrix}\right\}$ and $\dim E_{3}=1$. Comparison: geometric $=1<2=$ algebraic. The criterion fails, $A$ is NOT diagonalizable ✓. Concretely, there is only ONE eigendirection in a space of dimension 2: it is impossible to build a basis of eigenvectors, and therefore impossible to form an invertible matrix $P$.

$B=3I$ is also triangular, with eigenvalue 3 of algebraic multiplicity 2. But here $B-3I=\begin{pmatrix}0&0\\0&0\end{pmatrix}$, the ZERO matrix: the system $(B-3I)\vec{v}=\vec{0}$ imposes NO condition, every vector is a solution, so $E_{3}=\mathbb{R}^{2}$ and $\dim E_{3}=2$. Geometric $=2=$ algebraic: the criterion is satisfied and $B$ is diagonalizable. It is even diagonal already, with $P=I$. What changes between $A$ and $B$ is the off-diagonal entry: $B=3I$ stretches every direction uniformly, so that EVERY non-zero vector is an eigenvector; $A$, on the other hand, adds a SHEAR that leaves only one direction invariant. Two matrices with the same characteristic polynomial can therefore behave in opposite ways: the characteristic polynomial alone never decides diagonalizability, and that is the reason the geometric multiplicity exists.

Proof. Let $\lambda_{1},\ldots,\lambda_{n}$ be the $n$ distinct eigenvalues, and $\vec{v}_{1},\ldots,\vec{v}_{n}$ associated eigenvectors. The key point is that eigenvectors associated with DISTINCT eigenvalues are linearly independent. Here it is for two: suppose $c_{1}\vec{v}_{1}+c_{2}\vec{v}_{2}=\vec{0}$. Applying $A$: $c_{1}\lambda_{1}\vec{v}_{1}+c_{2}\lambda_{2}\vec{v}_{2}=\vec{0}$. Multiplying the first relation by $\lambda_{1}$ and subtracting: $c_{2}(\lambda_{2}-\lambda_{1})\vec{v}_{2}=\vec{0}$. Since $\lambda_{2}\neq\lambda_{1}$ and $\vec{v}_{2}\neq\vec{0}$, we get $c_{2}=0$, then $c_{1}=0$ ✓ (the general case is handled the same way by induction). We therefore have $n$ independent vectors in a space of dimension $n$: they form a basis of it, made entirely of eigenvectors, and $A$ is diagonalizable ✓. Application: $\begin{pmatrix}5&7\\0&-2\end{pmatrix}$ is triangular, its eigenvalues are 5 and $-2$, which are DISTINCT, two of them in a space of dimension 2. It is therefore diagonalizable, without computing a single eigenvector. Be careful not to reverse the implication: the condition is SUFFICIENT, not necessary, as $B=3I$ from part c) shows, diagonalizable with a double eigenvalue.