Let $A$ and $B$ be $n \times n$ matrices.

Is it always true that $\tr (A B) = \tr (A) \tr (B) $?

If it is true, prove it. If not, give a counterexample.

Solution.

There are many counterexamples.

For one, take
\[A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \text{ and } B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.\]

Then $\tr(A)=1, \tr(B)=1$, and hence $\tr(A) \tr(B) = 1$, while $\tr(AB) = 0$ as $AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

The post Does the Trace Commute with Matrix Multiplication? Is $\tr (A B) = \tr (A) \tr (B) $? first appeared on Problems in Mathematics.

Let $A$ be an $n \times n$ matrix.

Is it true that $\tr ( A^\trans ) = \tr(A)$? If it is true, prove it. If not, give a counterexample.

Solution.

The answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the diagonal entries of the transposed matrix are the same as the original matrix.

Putting together these observations yields the equality $\tr ( A^\trans ) = \tr(A)$.

Here is the more formal proof.

For $A = (a_{i j})_{1 \leq i, j \leq n}$, the transpose $A^{\trans}= (b_{i j})_{1 \leq i, j \leq n}$ is defined by $b_{i j} = a_{j i}$.

In particular, notice that $b_{i i} = a_{i i}$ for $1 \leq i \leq n$. And so,
\[ \tr(A^{\trans}) = \sum_{i=1}^n b_{i i} = \sum_{i=1}^n a_{i i} = \tr(A) . \]

The post Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix? first appeared on Problems in Mathematics.

Let $A$ and $B$ be $2\times 2$ matrices such that $(AB)^2=O$, where $O$ is the $2\times 2$ zero matrix.

Determine whether $(BA)^2$ must be $O$ as well. If so, prove it. If not, give a counter example.

Proof.

It is true that the matrix $(BA)^2$ must be the zero matrix as we shall prove now.

For notational convenience, let $C:=AB$.

As $C$ is a $2\times 2$ matrix, it satisfies the relation
\[C^2-\tr(C)C+\det(C)I=O \tag{*}\] by the Cayley-Hamilton theorem.
Here $I$ is the $2\times 2$ identity matrix.

We compute the determinant of $C$ as follows.
We have
\begin{align*}
\det(C)^2=\det(C^2)=\det((AB)^2)=\det(O)=0.
\end{align*}
Hence $\det(C)=0$.

Since $C^2=O$ and $\det(C)=0$, the Cayley-Hamilton relation (*) becomes
\[\tr(C)C=O.\] It follows that we have either $C=O$ or $\tr(C)=0$.
In either case, we have $\tr(C)=0$.

Note that
\begin{align*}
\det(BA)&=\det(AB)=\det(C)=0\\
\tr(BA)&=\tr(AB)=\tr(C)=0.
\end{align*}

Applying the Cayley-Hamilton theorem to the matrix $BA$, we obtain
\begin{align*}
O=(BA)^2-\tr(BA)\cdot BA+\det(BA)I=(BA)^2.
\end{align*}
Thus, we obtain $(BA)^2=O$ as claimed.

The post True or False: If $A, B$ are 2 by 2 Matrices such that $(AB)^2=O$, then $(BA)^2=O$ first appeared on Problems in Mathematics.

(a) Is the matrix $A=\begin{bmatrix}
1 & 2\\
0& 3
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
3 & 0\\
1& 2
\end{bmatrix}$?  

(b) Is the matrix $A=\begin{bmatrix}
0 & 1\\
5& 3
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
1 & 2\\
4& 3
\end{bmatrix}$? 

(c) Is the matrix $A=\begin{bmatrix}
-1 & 6\\
-2& 6
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
3 & 0\\
0& 2
\end{bmatrix}$? 

(d) Is the matrix $A=\begin{bmatrix}
-1 & 6\\
-2& 6
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
1 & 2\\
-1& 4
\end{bmatrix}$?

Solution.

(a) Is the matrix $A=\begin{bmatrix}
1 & 2\\
0& 3
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
3 & 0\\
1& 2
\end{bmatrix}$?

Recall that if $A$ and $B$ are similar, then their determinants are the same.
We compute
\begin{align*}
\det(A)=(1)(3)-(2)(0)=3 \text{ and } \det(B)=(3)(2)-(0)(1)=6.
\end{align*}
Thus, $\det(A)\neq \det(B)$, and hence $A$ and $B$ are not similar.

(b) Is the matrix $A=\begin{bmatrix}
0 & 1\\
5& 3
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
1 & 2\\
4& 3
\end{bmatrix}$?

It is straightforward to check that $\det(A)=-5=\det(B)$. Thus determinants does not help here.
We recall that if $A$ and $B$ are similar, then their traces are the same. (See Problem “Similar matrices have the same eigenvalues“.)
We compute
\begin{align*}
\tr(A)=0+3=3 \text{ and } \tr(B)=1+3=4,
\end{align*}
and thus $\tr(A)\neq\tr(B)$.
Hence $A$ and $B$ are not similar.

(c) Is the matrix $A=\begin{bmatrix}
-1 & 6\\
-2& 6
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
3 & 0\\
0& 2
\end{bmatrix}$?

We see that
\[\det(A)=6=\det(B) \text{ and } \tr(A)=5=\tr(B).\] Thus, the determinants and traces do not give any information about similarity.
The characteristic polynomial of $A$ is given by
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
-1-t & 6\\
-2& 6-t
\end{vmatrix}\\
&=(-1-t)(6-t)-(6)(-2)\\
&=t^2-5t+6.
\end{align*}
(Note that since we found the determinant and trace of $A$, we could have found the characteristic polynomial from the formula $p(t)=t^2-\tr(A)t+\det(A)$.)

Since $p(t)=(t-2)(t-3)$, the eigenvalue of $A$ are $2$ and $3$.
Since $A$ has two distinct eigenvalues, it is diagonalizable.
That is, there exists a nonsingular matrix $S$ such that
\[S^{-1}AS=\begin{bmatrix}
2 & 0\\
0& 3
\end{bmatrix}=B.\] Thus, $A$ and $B$ are similar.

(d) Is the matrix $A=\begin{bmatrix}
-1 & 6\\
-2& 6
\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}
1 & 2\\
-1& 4
\end{bmatrix}$?

We see that
\[\det(A)=6=\det(B) \text{ and } \tr(A)=5=\tr(B).\] It follows from the formula $p(t)=t^2-\tr(A)t+\det(A)$ (or just computing directly) that the characteristic polynomials of $A$ and $B$ are both
\[t^2-5t+6=(t-2)(t-3).\] Thus, the eigenvalues of $A$ and $B$ are $2, 3$. Hence both $A$ and $B$ are diagonalizable.
There exist nonsingular matrices $S$ and $P$ such that
\[S^{-1}AS=\begin{bmatrix}
2 & 0\\
0& 3
\end{bmatrix} \text{ and } P^{-1}BP=\begin{bmatrix} 2 & 0\\
0& 3
\end{bmatrix}. \] So we have $S^{-1}AS=P^{-1}BP$, and hence
\[PS^{-1}ASP^{-1}=B.\] Putting $U=SP^{-1}$, we have
\[U^{-1}AU=B.\] (Since the product of invertible matrices is invertible, the matrix $U$ is invertible.)
Therefore $A$ and $B$ are similar.

Related Question.

For more problems about similar matrices, check out the following posts:

• Problems and solutions about similar matrices
• A matrix similar to a diagonalizable matrix is also diagonalizable

The post Determine Whether Given Matrices are Similar first appeared on Problems in Mathematics.

Prove that if $A$ and $B$ are similar matrices, then their determinants are the same.

Proof.

Suppose that $A$ and $B$ are similar. Then there exists a nonsingular matrix $S$ such that
\[S^{-1}AS=B\] by definition.
Then we have
\begin{align*}
&\det(B)\\
&=\det(S^{-1}AS)\\
&=\det(S)^{-1}\det(A)\det(S) \\
& \text{(by multiplicative properties of determinants)}\\
&=\det(A) \\
&\text{(since determinants are just numbers, hence commutative).}
\end{align*}

Thus, we obtain $\det(A)=\det(B)$ as required.

Related Question.

More generally, we can prove that if $A$ and $B$ are similar, then their characteristic polynomials are the same.
From this, we also can deduce that the determinants of $A$ and $B$ are the same as well as their traces are the same.

For a proof, see the post “Similar matrices have the same eigenvalues“.

The post If Two Matrices are Similar, then their Determinants are the Same first appeared on Problems in Mathematics.

(a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$.
Find $\det(A)$.

(b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$.

(c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$?

(Harvard University, Linear Algebra Exam Problem)

Solution.

For (a) and (b), we give two solutions. The first one does not use the knowledge of eigenvalues, and the second one uses eigenvalues.

Solution 1 of (a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$.

Let
\[A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}.\] Then we have
\begin{align*}
A^2=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}=\begin{bmatrix}
a^2+bc & ab+bd\\
ac+cd& bc+d^2
\end{bmatrix}.
\end{align*}
Since $\tr(A^2)=5$ and $\tr(A)=3$, we obtain
\begin{align*}
5&=\tr(A^2)=(a^2+bc)+(bc+d^2)=a^2+2bc+d^2 \text{ and }\\
3&=\tr(A)=a+d.
\end{align*}
We find the determinant $\det(A)=ad-bc$ as follows.
We have
\begin{align*}
\det(A)&=ad-bc=\frac{1}{2}\left(\, (a+d)^2-(a^2+2bc+d^2) \,\right)\\
&=\frac{1}{2}(3^2-5)=2.
\end{align*}
Thus, we obtain $\det(A)=2$.

Solution 2 of (a)

Let $\lambda_1$ and $\lambda_2$ be eigenvalues of $A$.
Then we have
\begin{align*}
3=\tr(A)=\lambda_1+\lambda_2 \text{ and }\\
5=\tr(A^2)=\lambda_1^2+\lambda_2^2.
\end{align*}
Here we used two facts.
The first one is that the trace of a matrix is the sum of all eigenvalues of the matrix.
The second one is that $\lambda^2$ is an eigenvalue of $A^2$ if $\lambda$ is an eigenvalue of $A$, and these are all the eigenvalues of $A^2$.

Since the determinant of $A$ is the product of eigenvalues of $A$, we have
\begin{align*}
\det(A)&=\lambda_1 \lambda_2\\
&=\frac{1}{2}\left(\, (\lambda_1+\lambda_2)^2-(\lambda_1^2+\lambda_2^2) \,\right)\\
&=\frac{1}{2}(3^2-5)\\
&=2.
\end{align*}
Hence we have $\det(A)=2$.

Solution 1 of (b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$.

Since two columns are parallel, we can write $A$ as
\[A=\begin{bmatrix}
a & ra\\
c& rc
\end{bmatrix}.\] Then we have
\begin{align*}
5=\tr(A)=a+rc.
\end{align*}
We use the formula in Solution 1 of (a) for $\tr(A^2)$ with $b=ra$ and $d=rc$, and we compute
\begin{align*}
\tr(A^2)&=a^2+2(ra)+(rc)^2\\
&=(a+rc)^2\\
&=5^2=25.
\end{align*}
Thus, we find $\tr(A^2)=25$.

Solution 2 of (b)

Since two columns are parallel, the matrix $A$ is singular. Hence $A$ has an eigenvalue $0$.
Since the sum of all the eigenvalues is $\tr(A)=5$, we see that $0$ and $5$ are eigenvalues of $A$.
It follows that $0$ and $25$ are eigenvalues of $A^2$. Hence
\[\tr(A^2)=0+25=25.\]

Solution of (c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$?

The product of eigenvalues of $A$ is the determinant $\det(A)=5$.
Since eigenvalues are positive integers, it follows that $1$ and $5$ are eigenvalues of $A$.
It follows that
\[\tr(A)=1+5=6.\] Click here if solved 71

The post Trace, Determinant, and Eigenvalue (Harvard University Exam Problem) first appeared on Problems in Mathematics.

Let $A, B$ be complex $2\times 2$ matrices satisfying the relation
\[A=AB-BA.\]

Prove that $A^2=O$, where $O$ is the $2\times 2$ zero matrix.

Hint.

1. Find the trace of $A$.
2. Use the Cayley-Hamilton theorem

Proof.

We first calculate the trace of the matrix $A$ as follows. We have
\begin{align*}
\tr(A)&=\tr(AB-BA)\\
&=\tr(AB)-\tr(BA)\\
&=\tr(AB)-\tr(AB)=0.
\end{align*}

Thus $\tr(A)=0$ and it follows from the Cayley-Hamilton theorem (see below) for the $2\times 2$ matrix $A$ that
\begin{align*}
O&=A^2-\tr(A)A+\det(A)I\\
&=A^2+\det(A)I,
\end{align*}
where $I$ is the $2\times 2$ identity matrix.

Thus, we obtain
\[A^2=-\det(A)I. \tag{*}\]

Next, we compute $A^2$ in two ways.
We have
\begin{align*}
A^2=A(AB-BA)=A^2B-ABA
\end{align*}
and
\begin{align*}
A^2=(AB-BA)A=ABA-BA^2.
\end{align*}
Adding these two, we have
\begin{align*}
2A^2&=A^2B-BA^2\\
& \stackrel{(*)}{=} (-\det(A)I)B-B(-\det(A)I)\\
&=-\det(A)B+\det(A)B=O.
\end{align*}

As a result, we obtain $A^2=O$. This completes the proof.

The Cayley-Hamilton theorem for a $2\times 2$ matrix

Let us add the proof of the fact we used in the proof about the Cayley-Hamilton theorem.
Let $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ be a $2\times 2$ matrix.

Then its characteristic polynomial is
\begin{align*}
p(x)&=\det(A-xI)\\
&=\begin{vmatrix}
a-x & b\\
c& d-x
\end{vmatrix}\\
&=(a-x)(d-x)-bc\\
&=x^2-(a+d)x+ad-bc\\
&=x^2-\tr(A)x+\det(A),
\end{align*}
since $\tr(A)=a+d$ and $\det(A)=ad-bc$.

The Cayley-Hamilton theorem says that the matrix $A$ satisfies its characteristic equation $p(x)=0$.
Namely we have
\[A^2-\tr(A)A+\det(A)I=O.\] This is the equality we used in the proof.

Variation

As a variation of this problem, consider the following problem.

Let $A, B$ be $2\times 2$ matrices satisfying $A=AB-BA$.
Then prove that $\det(A)=0$.

The post If 2 by 2 Matrices Satisfy $A=AB-BA$, then $A^2$ is Zero Matrix first appeared on Problems in Mathematics.

Let $I$ be the $n\times n$ identity matrix, where $n$ is a positive integer. Prove that there are no $n\times n$ matrices $X$ and $Y$ such that
\[XY-YX=I.\]  

Hint.

Suppose that such matrices exist and consider the trace of the matrix $XY-YX$.

Recall that the trace of a square matrix $A$ is the sum of the diagonal entries of $A$.

Proof.

We use the following basic properties of the trace of matrices.
Let $A, B$ be $n\times n$ matrices and let $c$ be a scalar.

1. $\tr(A+B)=\tr(A)+\tr(B)$.
2. $\tr(cA)=c\tr(A)$.
3. $\tr(AB)=\tr(BA)$.

Seeking a contradiction, we assume that there are matrices $X$ and $Y$ such that $XY-YX=I$.

Then we take the trace of both sides and obtain
\begin{align*}
n&=\tr(I)\\
&=\tr(XY-YX)\\
&=\tr(XY)-\tr(YX) \qquad \text{ (by property (1), (2) of the trace)}\\
&=\tr(XY)-\tr(YX) \qquad \text{ (by property (3) of the trace)}\\
&=0.
\end{align*}

Since $n$ is a positive integer, this is a contradiction.
Therefore, such matrices $X, Y$ do not exist.

The post Matrix $XY-YX$ Never Be the Identity Matrix first appeared on Problems in Mathematics.

Let $V$ be the set of all $n \times n$ diagonal matrices whose traces are zero.
That is,

\begin{equation*}
V:=\left\{ A=\begin{bmatrix}
a_{11} & 0 & \dots & 0 \\
0 &a_{22} & \dots & 0 \\
0 & 0 & \ddots & \vdots \\
0 & 0 & \dots & a_{nn}
\end{bmatrix} \quad \middle| \quad
\begin{array}{l}
a_{11}, \dots, a_{nn} \in \C,\\
\tr(A)=0 \\
\end{array}
\right\}
\end{equation*}

Let $E_{ij}$ denote the $n \times n$ matrix whose $(i,j)$-entry is $1$ and zero elsewhere.

(a) Show that $V$ is a subspace of the vector space $M_n$ over $\C$ of all $n\times n$ matrices. (You may assume without a proof that $M_n$ is a vector space.)

(b) Show that matrices
\[E_{11}-E_{22}, \, E_{22}-E_{33}, \, \dots,\, E_{n-1\, n-1}-E_{nn}\] are a basis for the vector space $V$.

(c) Find the dimension of $V$.

Hint (Subspace Criteria)

For (a), use the criteria for a subset to be a subspace.

1. The zero vector $\mathbf{0} \in M_n$ is in $V$.
2. For $\mathbf{u}, \mathbf{v}\in V$, the sum $\mathbf{u}+\mathbf{v}\in V$.
3. For $\mathbf{v}\in V, c\in \C$, the scalar product $c\mathbf{v}\in V$.

Proof.

(a) $V$ is a subspace of the vector space $M_n$ over $\C$

We check the following criteria for a subset to be a subspace.

1. The zero vector $\mathbf{0} \in M_n$ is in $V$.
2. For $\mathbf{u}, \mathbf{v}\in V$, the sum $\mathbf{u}+\mathbf{v}\in V$.
3. For $\mathbf{v}\in V, c\in \C$, the scalar product $c\mathbf{v}\in V$.

For (1), note that the zero vector $\mathbf{0}\in M_n$ is the $n \times n$ zero matrix. Thus it is clearly in $V$.

To check (2), take $\mathbf{u}, \mathbf{v}\in V$. Then $\mathbf{u}, \mathbf{v}$ are diagonal matrices and $\tr(\mathbf{u})=0$, $\tr(\mathbf{v})=0$.
Thus the sum $\mathbf{u}+\mathbf{v}$ is a diagonal matrix and the trace is
\[\tr(\mathbf{u}+\mathbf{v})=\tr(\mathbf{u})+\tr(\mathbf{v})=0+0=0.\] Hence the sum $\mathbf{u}+\mathbf{v} \in V$.

Finally, to prove (3) we take $\mathbf{v}\in V, c\in \C$.
Then $\mathbf{v}$ is a diagonal matrix and $\tr(\mathbf{v})=0$.
Thus $c\mathbf{v}$ is also a diagonal matrix and the trace is
\[\tr(c\mathbf{v})=c\tr(\mathbf{v})=c\cdot 0=0.\] Hence $c\mathbf{v} \in V$.
Therefore the subspace criteria (1)-(3) hold. Thus $V$ is a subspace of the vector space $M_n$.

(b) A basis for the vector space $V$.

Note that for each $1 \leq i \leq n-1$, the matrix $E_{ii}-E_{i+1\, i+1}$ is a diagonal matrix and the trace is
\[\tr(E_{ii}-E_{i+1\, i+1})=\tr(E_{ii})-\tr(E_{i+1\,i+1})=1-1=0.\] Hence the matrix $E_{ii}-E_{i+1\, i+1} \in V$.

We show that the matrices $E_{ii}-E_{i+1\, i+1}$ span $V$.
Let $A=(a_{ij})\in V$. We want to solve the following linear combination.
\begin{align*}
A=c_1(E_{11}-E_{22})+c_2(E_{22}-E_{33})+\cdots+c_{n-1}( E_{n-1\, n-1}-E_{nn}).
\end{align*}
Rearranging this, we obtain
\begin{align*}
A=c_1E_{11}+(c_2-c_1)E_{22}+\cdots+(c_{n-1}-c_{n-2}) E_{n-1\, n-1}-c_{n-1}E_{nn}.
\end{align*}

Namely, in the matrix form we have
\[\begin{bmatrix}
a_{11} & 0 & \dots & 0 \\
0 &a_{22} & \dots & 0 \\
\vdots & \ddots & \ddots & \ddots \\
0 & 0 & \dots & a_{nn}
\end{bmatrix}
=
\begin{bmatrix}
c_1 & 0 & \dots & \dots &0 \\
0 & c_2-c_1 & \dots & \dots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \dots & 0 & c_{n-1}-c_{n-2} & 0 \\
0 & \dots & \dots & 0 & -c_{n-1}
\end{bmatrix}.\] Comparing entries, we have the system of $n$ equations in $n-1$ unknowns $c_1,\dots, c_{n-1}$
\begin{align*}
a_{11}&=c_1\\
a_{22}&=c_2-c_1\\
a_{33}&=c_3-c_2\\
\vdots\\
a_{n-1 n-1}&=c_{n-1}-c_{n-2}\\
a_{nn}&=-c_{n-1}.
\end{align*}

We solve this system as follows.
The matrix form of this system is
\[\begin{bmatrix}
1 & & & & \\
-1 & 1 & & & \\
& -1 & 1 & & \\
& & \ddots & \ddots & \\
& & & -1& 1\\
& & & & -1
\end{bmatrix}\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_{n-1}
\end{bmatrix}=\begin{bmatrix}
a_{11} \\
a_{22} \\
\vdots \\
a_{nn}
\end{bmatrix}.\] (The empty entries are all zero.)

The augmented matrix is
\[\left[\begin{array}{rrrrr|r}
1 & & & & & a_{11}\\
-1 & 1 & & & &a_{22} \\
& -1 & 1 & & &a_{33}\\
& & \ddots & \ddots & & \vdots \\
& & & -1& 1 &a_{n-1 n-1}\\
& & & & -1 &a_{n n}
\end{array} \right].\] Then we reduce this matrix as follows.
Add the first row to the second row, then add the second row to the third row.
Repeating this process we get
\[\left[\begin{array}{rrrrr|r}
1 & & & & & a_{11}\\
& 1 & & & &a_{11}+a_{22} \\
& & 1 & & &a_{11}+a_{22}+a_{33}\\
& & \ddots & \ddots & & \vdots \\
& & & & 1 &a_{11}+a_{22}+\dots +a_{n-1 n-1}\\
& & & & 0 &a_{11}+a_{22}+\dots +a_{n n}
\end{array} \right].\]

Note that since $A \in V$, $\tr(A)=a_{11}+a_{22}+\cdots a_{nn}=0$, thus the right-bottom entry ($(n,n)$-entry) of the augmented matrix is $0$, hence the system is consistent and it has the unique solution
\begin{align*}
c_{1} &= a_{11} \\
c_{2} &=a_{11}+a_{22}\\
c_3 &=a_{11}+a_{22}+a_{33}\\
\vdots \\
c_{n-1}&=a_{11}+a_{22}+\cdots+a_{n-1 n-1}.\tag{*}
\end{align*}
Therefore the matrices $E_{11}-E_{22}, \, E_{22}-E_{33}, \, \dots,\, E_{n-1\, n-1}-E_{nn}$ span $V$.

To show that these matrices are linearly independent, we can reuse the above computation.
Suppose we have a linear combination
\[\mathbf{0}=c_1(E_{11}-E_{22})+c_2(E_{22}-E_{33})+\cdots+c_{n-1}( E_{n-1\, n-1}-E_{nn}).\] Here $\mathbf{0}$ is the $n \times n$ zero matrix.
Since $a_{ii}=0$ in the above computation, we see from (*) that $c_1=c_2=\dots=c_{n-1}=0$, hence the matrices $E_{ii}-E_{i+1 i+1}$ are linearly independent.
Therefore those matrices are a basis of $V$.

(c) The dimension of $V$.

From part (b), the $n-1$ matrices
\[E_{11}-E_{22}, \, E_{22}-E_{33}, \, \dots,\, E_{n-1\, n-1}-E_{nn}\] are a basis of $V$. Thus the dimension of $V$ is $n-1$.

The post The Vector Space Consisting of All Traceless Diagonal Matrices first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.
Show that

(1) $$\det(A)=\prod_{i=1}^n \lambda_i$$

(2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$

Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix $A$.

Namely, prove that (1) the determinant of $A$ is the product of its eigenvalues, and (2) the trace of $A$ is the sum of the eigenvalues.
We give two different proofs.

Plan 1.

1. Use the definition of eigenvalues (the characteristic polynomial).
2. Compare coefficients.

Plan 2.

1. Make $A$ upper triangular matrix or in the Jordan normal/canonical form.
2. Use the property of determinants and traces.

Proof. [Method 1]

(1) Recall that eigenvalues are roots of the characteristic polynomial $p(\lambda)=\det(A-\lambda I_n)$.
It follows that we have
\begin{align*}
&\det(A-\lambda I_n) \\
&=\begin{vmatrix}
a_{1 1}- \lambda & a_{1 2} & \cdots & a_{1,n} \\
a_{2 1} & a_{2 2} -\lambda & \cdots & a_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{m 2} & \cdots & a_{n n}-\lambda
\end{vmatrix} =\prod_{i=1}^n (\lambda_i-\lambda). \tag{*}
\end{align*}

Letting $\lambda=0$, we see that $\det(A)=\prod_{i=1}^n \lambda_i$ and this completes the proof of part (a).

(2) Compare the coefficients of $\lambda^{n-1}$ of the both sides of (*).
The coefficient of $\lambda^{n-1}$ of the determinant on the left side of (*) is

$$(-1)^{n-1}(a_{11}+a_{22}+\cdots a_{n n})=(-1)^{n-1}\tr(A).$$
The coefficient of $\lambda^{n-1}$ of the determinant on the right side of (*) is
$$(-1)^{n-1}\sum_{i=1}^n \lambda_i.$$
Thus we have $\tr(A)=\sum_{i=1}^n \lambda_i$.

Proof. [Method 2]

Observe that there exists an $n \times n$ invertible matrix $P$ such that
\[P^{-1} A P= \begin{bmatrix}
\lambda_1 & * & \cdots & * \\
0 & \lambda_2 & \cdots & * \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n \tag{**}
\end{bmatrix}.
\] This is an upper triangular matrix and diagonal entries are eigenvalues.
(If this is not familiar to you, then study a “triangularizable matrix” or “Jordan normal/canonical form”.)

(1) Since the determinant of an upper triangular matrix is the product of diagonal entries, we have
\begin{align*}
\prod_{i=1}^n \lambda_i & =\det(P^{-1} A P)=\det(P^{-1}) \det(A) \det(P) \\
&= \det(P)^{-1}\det(A) \det(P)=\det(A),
\end{align*}
where we used the multiplicative property of the determinant.

(2) We take the trace of both sides of (**) and get
\begin{align*}
\sum_{i=1}^n \lambda_i =\tr(P^{-1}AP) =\tr(A).
\end{align*}
(Here for the last equality we used the property of the trace that $\tr(AB)=\tr(BA)$ for any $n\times n$ matrices $A$ and $B$.)
Thus we obtained the result $\tr(A)=\sum_{i=1}^n \lambda_i$.

Comment.

The proof of (1) in the first method is simple, but that of (2) requires a bit observation, especially when we find the coefficient of the left-hand side.

The proof of (2) in the second method is simpler although you need to know about the Jordan normal/canonical form.

These two formulas relate the determinant and the trace, and the eigenvalue of a matrix in a very simple way.

The post Determinant/Trace and Eigenvalues of a Matrix first appeared on Problems in Mathematics.