Let $\mathrm{P}_2$ denote the vector space of polynomials of degree $2$ or less, and let $T : \mathrm{P}_2 \rightarrow \mathrm{P}_2$ be the derivative linear transformation, defined by
\[ T( ax^2 + bx + c ) = 2ax + b . \]

Is $T$ diagonalizable? If so, find a diagonal matrix which represents $T$. If not, explain why not.

Solution.

The standard basis of the vector space $\mathrm{P}_2$ is the set $B = \{ 1 , x , x^2 \}$. The matrix representing $T$ with respect to this basis is
\[ [T]_B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} . \]

The characteristic polynomial of this matrix is
\[ \det ( [T]_B – \lambda I ) = \begin{vmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 2 \\ 0 & 0 & -\lambda \end{vmatrix} = \, – \lambda^3 . \] We see that the only eigenvalue of $T$ is $0$ with algebraic multiplicity $3$.

On the other hand, a polynomial $f(x)$ satisfies $T(f)(x) = 0$ if and only if $f(x) = c$ is a constant. The null space of $T$ is spanned by the single constant polynomial $\mathbb{1}(x) = 1$, and thus is one-dimensional. This means that the geometric multiplicity of the eigenvalue $0$ is only $1$.

Because the geometric multiplicity of $0$ is less than the algebraic multiplicity, the map $T$ is defective, and thus not diagonalizable.

The post Is the Derivative Linear Transformation Diagonalizable? first appeared on Problems in Mathematics.

Suppose that the $2 \times 2$ matrix $A$ has eigenvalues $4$ and $-2$. For each integer $n \geq 1$, there are real numbers $b_n , c_n$ which satisfy the relation
\[ A^{n} = b_n A + c_n I , \] where $I$ is the identity matrix.

Find $b_n$ and $c_n$ for $2 \leq n \leq 5$, and then find a recursive relationship to find $b_n, c_n$ for every $n \geq 1$.

Solution.

Because the eigenvalues of $A$ are $4$ and $-2$, its characteristic polynomial must be
\[ p(\lambda) = (\lambda – 4) ( \lambda + 2) = \lambda^2 – 2 \lambda – 8 . \]

The Cayley-Hamilton Theorem tells us that $p(A) = 0$, the zero matrix. Rearranging terms, we find that
\[ A^2 = 2 A + 8 I . \] So $b_2=2$ and $c_2=8$.

With this relationship, we can reduce the higher powers of $A$:
\begin{align*}
A^3 &= A^2 A= (2A + 8 I)A = 2A^2 + 8 A\\
& = 2(2A+8I)+8A= 12 A + 16 I
\end{align*}
\begin{align*}
A^4 &= A^3 A=(12 A + 16 I)A = 12A^2+16A\\
&=12(2A+8I)+16A = 40 A + 96 I
\end{align*}
\begin{align*}
A^5 &=A^4A= (40 A + 96 I)A = 40 A^2 + 96 A \\
&=40(2A+8I)+96A = 176 A + 320 I
\end{align*}
Hence, we have $b_3=12, c_3=16, b_4=40, c_4=96, b_5=176$, and $c_5=320$.

To find the recursive relationship, suppose we know that $A^n = b_n A + c_n I$. Then
\begin{align*}
A^{n+1} &= A^{n} A = ( b_n A + c_n I ) A = b_n A^2 + c_n A\\
&=b_n(2A+8I)+c_n A = (2 b_n + c_n) A + 8 b_n I.
\end{align*}
This gives the recursive relationships
\[ b_{n+1} = 2 b_n + c_n , \qquad c_{n+1} = 8 b_n . \]

Using the work above, you can quickly verify this for $1 \leq n \leq 4$.

The post A Recursive Relationship for a Power of a Matrix first appeared on Problems in Mathematics.

Let $A$ and $B$ be square matrices such that they commute each other: $AB=BA$.
Assume that $A-B$ is a nilpotent matrix.

Then prove that the eigenvalues of $A$ and $B$ are the same.

Proof.

Let $N:=A-B$. By assumption, the matrix $N$ is nilpotent.
This means that there exists a positive integer $n$ such that $N^n$ is the zero matrix $O$.

Let $\lambda$ be an eigenvalue of $B$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$. That is, we have $B\mathbf{v}=\lambda \mathbf{v}$ and $\mathbf{v}\neq \mathbf{0}$.
We prove that $\lambda$ is also an eigenvalue of $A$.

Note that since $A$ and $B$ commute each other, it follows that the matrices $N$ and $B-\lambda I$ commute each other as well.
Then we compute
\begin{align*}
(A-\lambda I)^n&=(N+B-\lambda I)^n\\
&=\sum_{i=0}^n \begin{pmatrix}
n \\
i
\end{pmatrix}
N^i(B-\lambda I)^{n-i},
\end{align*}
where the second equality follows by the binomial expansion. (Note that the binomial expansion is true for matrices commuting each other.)
Then we have
\begin{align*}
(A-\lambda I)^n \mathbf{v}&=\sum_{i=0}^{n-1} \begin{pmatrix}
n \\
i
\end{pmatrix}
N^i(B-\lambda I)^{n-i}\mathbf{v}+N^n\mathbf{v}=\mathbf{0}
\end{align*}
since $(B-\lambda I)\mathbf{v}=\mathbf{0}$ and $N^n=O$.

This implies that there exists an integer $k$, $0\leq k \leq n-1$ such that
\[\mathbf{u}:=(A-\lambda I)^k\mathbf{v}\neq \mathbf{0} \text{ and } (A-\lambda I)^{k+1}\mathbf{v}=\mathbf{0}.\]

It yields that $(A-\lambda I)\mathbf{u}=\mathbf{0}$ and $\mathbf{u}\neq \mathbf{0}$, or equivalently $A\mathbf{u}=\lambda \mathbf{u}$.
Hence $\lambda$ is an eigenvalue of $A$.

This proves that each eigenvalue of $B$ is an eigenvalue of $A$.
Note that if $A-B$ is nilpotent, then $B-A$ is also nilpotent.
Thus, switching the roles of $A$ and $B$, we also see that each eigenvalue of $A$ is an eigenvalue of $B$.
Therefore, the eigenvalues of $A$ and $B$ are the same.

The post Commuting Matrices $AB=BA$ such that $A-B$ is Nilpotent Have the Same Eigenvalues first appeared on Problems in Mathematics.

Let $I$ be the $2\times 2$ identity matrix.
Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$.

Proof.

Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies
\[ABA^{-1}=-B. \tag{*}\] Taking the trace, we have
\[-\tr(B)=\tr(-B)=\tr(ABA^{-1})=tr(BAA^{-1})=\tr(B),\] hence the trace $\tr(B)=0$.
Thus, the characteristic polynomial of $B$ is
\[x^2-\tr(B)x+\det(B)=x^2+1.\] Hence the eigenvalues of $B$ are $\pm i$.

Note that the matrix $\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}$ has also eigenvalues $\pm i$.
Thus this matrix is similar to the matrix $B$ as both matrices are similar to the diagonal matrix $\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}$.
Let $P$ be a nonsingular matrix such that
\[B’:=P^{-1}BP=\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}.\] Let $A’=P^{-1}AP$.

The relation (*) is equivalent to $AB=-BA$.
Using this we have
\begin{align*}
A’B’&=(P^{-1}AP)(P^{-1}BP)=P^{-1}(AB)P\\
&=P^{-1}(-BA)P=-(P^{-1}BP)(P^{-1}AP)=-B’A’.
\end{align*}

Let $A’=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$.
Then $A’B’=-B’A’$ gives
\begin{align*}
\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}
\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}
=-\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}\\[6pt] \Leftrightarrow
\begin{bmatrix}
b & -a\\
d& -c
\end{bmatrix}=\begin{bmatrix}
c & d\\
-a& -b
\end{bmatrix}.
\end{align*}
Hence we obtain $d=-a$ and $c=b$.

Then
\begin{align*}
1=\det(A)=\det(PA’P^{-1})=\det(A’)=\begin{vmatrix}
a & b\\
b& -a
\end{vmatrix}=-a^2-b^2,
\end{align*}
which is impossible.
Therefore, the matrix $-I$ cannot be written as a commutator $[A, B]$ for any $2\times 2$ matrices $A, B$ with determinant $1$.

The post An Example of a Matrix that Cannot Be a Commutator first appeared on Problems in Mathematics.

Answer the following questions regarding eigenvalues of a real matrix.

(a) True or False. If each entry of an $n \times n$ matrix $A$ is a real number, then the eigenvalues of $A$ are all real numbers.
(b) Find the eigenvalues of the matrix
\[B=\begin{bmatrix}
-2 & -1\\
5& 2
\end{bmatrix}.\]

(The Ohio State University, Linear Algebra Exam)

Hint.

Consider a $2\times 2$ matrix.
Then the eigenvalues are solutions of a quadratic polynomial.

Does a quadratic polynomial always have real solutions?

Solution.

(a) True or False. If each entry of an $n \times n$ matrix $A$ is a real number, then the eigenvalues of $A$ are all real numbers.

 False. In general, a real matrix can have a complex number eigenvalue. In fact, the part (b) gives an example of such a matrix.

(b) Find the eigenvalues of the matrix

 The characteristic polynomial for $B$ is
\[ \det(B-tI)=\begin{bmatrix}
-2-t & -1\\
5& 2-t
\end{bmatrix}=t^2+1.\]

The eigenvalues are the solutions of the characteristic polynomial. Thus solving $t^2+1=0$, we obtain eigenvalues $\pm i$, where $i=\sqrt{-1}$.
Thus the eigenvalue of a real matrix $B$ is pure imaginary numbers $\pm i$.

The post True or False: Eigenvalues of a Real Matrix Are Real Numbers first appeared on Problems in Mathematics.