GATE CSE 2022 | Question: 10

Question

Consider the following two statements with respect to the matrices $\textit{A}_{m \times n}, \textit{B}_{n \times m}, \textit{C}_{n \times n}$ and $ \textit{D}_{n \times n}.$

Statement $1: tr \text{(AB)} = tr \text{(BA)}$

Statement $2: tr \text{(CD)} = tr \text{(DC)}$

where $tr()$ represents the trace of a matrix. Which one of the following holds?

• A. Statement $1$ is correct and Statement $2$ is wrong.
• B. Statement $1$ is wrong and Statement $2$ is correct.
• C. Both Statement $1$ and Statement $2$ are correct.
• D. Both Statement $1$ and Statement $2$ are wrong.

Comments

Link to such theorems. Might be helpful.

Trace (linear algebra) From Wikipedia, the free encyclopedia

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Statement 1: $\operatorname{tr}(A B)=\operatorname{tr}(B A)$

Trace: sum of eigen value

$$

A_{m \times n} \&B_{n \times m}=A B_{m \times m}

$$

$$B_{n \times m} \& A_{m \times n}=B A_{n \times n}$$

So, whatever size of matrix $A B$ and $B A$.

$\Rightarrow$ Non-zero eigen value of $A B \& B A$ are same and other eigen value are zero.

example:

suppose. $A B_{4 \times 4}$ has 4 eigen value

$$

\left(\lambda_1=2, \lambda_2=1, \lambda_3=7, \lambda_4=8\right)

$$

and $\mathrm{BA}_{7 \times 7}$ has 7 eigen value

$\Rightarrow$ out of 7,4 eigen value of $B A$ must same cs $A B$. and rest are zero.

$$

\begin{aligned}

& \left.\lambda_1=2\right] \operatorname{}\;\;\;\;\; \lambda_5=0 \\

& \left.\begin{array}{l}

\lambda_2=1 \\

\lambda_3=7

\end{array}\right\} \begin{array}{ll}

\text { } & \lambda_6=0 \\

& \lambda_7=0

\end{array} \\

& \lambda_4=8 \\

&

\end{aligned}

$$

Statement 2: $\operatorname{tr}(C D)=\operatorname{tr}(D C)$

it's very easy Both matrices have the same non-zero eigenvalues and their sum is also the same.

Both statements are true.

option C Correct 

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Answer 1

Answer C

Please check this video: https://www.youtube.com/watch?v=9hz7m6B_LW8&list=PLIPZ2_p3RNHhGLQ1ZT37KLpBMAD90CM4_&index=39

Trace of a matrix and sum of eigen values of matrix are same.

$AB$ and $BA$ has same nonzero eigen values (Proof below) hence their sum of eigen values i.e.  trace should also be same.

 

Proof: Let $\lambda$ be a nonzero eigenvalue of $AB$, and $x$ be the eigenvector of $AB$. We need to find the eigenvalue and eigenvector of $BA$.

\[
ABx = \lambda x \quad \text{(given, where $\lambda \neq 0$)}
\]

Since $\mathbf{\lambda}$ is an eigenvalue of $AB$, it follows that: \( \lambda\quad \text{is an eigenvalue of $BA$}\)

To find the eigenvector of $BA$, we multiply both sides of the equation by matrix \( B \) on the left:
\[
B(ABx) = B(\lambda x)
\]

On the right-hand side, since \( \lambda \) is a scalar, we can pull it outside of the multiplication and we use the associative property of matrix multiplication to group the matrices on the left-hand side::
\[
B(ABx) = \lambda (Bx)
\]

\[
(BA)(Bx) = \lambda (Bx)
\]

\[
BA(Bx) = \lambda (Bx)
\]

Thus, Eigenvector of $BA$ is $Bx$ and Eigenvalue of $BA$ is $\lambda$.

( Nonzero eigenvalues of $AB$ and $BA$ are the same )

Lets take a look at this example:

Given: Order of matrix $A$ is $5 \times 10$ and Order of matrix $B$ is $10 \times 5$

Order of \( AB : 5 \times 5 \quad \rightarrow \quad \text{Eigenvalues of $AB$: } 1, 2, 3, 4, 5 \)

Order of \(BA : 10 \times 10 \quad \rightarrow \quad \text{Eigenvalues of $BA$: } 1, 2, 3, 4, 5, 0, 0, 0, 0, 0\)

 

 

Comments

great explanation

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Suppose A is 3×2 matrix and B is 2×3 matix then AB is 3×3 matix with eigen values 6,7,8(assume)

And BA is 2×2 matrix, as we know it shares non zero eigen values so BA eigen values has to be 6,7,8 which is not possible due to dimension, so isn't is false?
​

 

---

@Anmol Chaurasia you are right but this situtation is impossible.

A $3 \times 3$ matrix $AB$ formed by a $3 \times 2$ matrix $A$ and a $2 \times 3$ matrix $B$ can never have three non-zero eigenvalues like $\{6, 7, 8\}$

we know,

$$\text{rank}(AB) \le \min(\text{rank}(A), \text{rank}(B))$$

• Since $A$ is $3 \times 2$, its maximum possible rank is $2$.

• Since $B$ is $2 \times 3$, its maximum possible rank is $2$.

• Therefore, the rank of the $3 \times 3$ matrix $AB$ can be at most 2.

coming back to the scenerio,

Since $\text{rank}(AB) \le 2$, the $3 \times 3$ matrix $AB$ must have at least one eigenvalue equal to 0

If $AB$ has valid eigenvalues under these dimensions, they must look something like $\{6, 7, 0\}$

Since $BA$ shares the exact same non-zero eigenvalues as $AB$, its eigenvalues will be precisely $\{6, 7\}$.

Answer 2

Statement -1 is evident from the definition of matrix product 

 

Statement-2 follows from statement-1, just that now both are square matrices

 

So, both are TRUE 

option C

Comments

This directly comes from frobenius natural inner product of real or complex matrices over the vector space.

Answer 3

Option (C)  : Both statement 1 and statement 2 are correct .

 

Answer 4

Statement 1: $\operatorname{tr}(A B)=\operatorname{tr}(B A)$

Trace: sum of eigen value

$$

A_{m \times n} \&B_{n \times m}=A B_{m \times m}

$$

$$B_{n \times m} \& A_{m \times n}=B A_{n \times n}$$

So, whatever size of matrix $A B$ and $B A$.

$\Rightarrow$ Non-zero eigen value of $A B \& B A$ are same and other eigen value are zero.

example:

suppose. $A B_{4 \times 4}$ has 4 eigen value

$$

\left(\lambda_1=2, \lambda_2=1, \lambda_3=7, \lambda_4=8\right)

$$

and $\mathrm{BA}_{7 \times 7}$ has 7 eigen value

$\Rightarrow$ out of 7,4 eigen value of $B A$ must same cs $A B$. and rest are zero.

$$

\begin{aligned}

& \left.\lambda_1=2\right] \operatorname{}\;\;\;\;\; \lambda_5=0 \\

& \left.\begin{array}{l}

\lambda_2=1 \\

\lambda_3=7

\end{array}\right\} \begin{array}{ll}

\text { } & \lambda_6=0 \\

& \lambda_7=0

\end{array} \\

& \lambda_4=8 \\

&

\end{aligned}

$$

Statement 2: $\operatorname{tr}(C D)=\operatorname{tr}(D C)$

it's very easy Both matrices have the same non-zero eigenvalues and their sum is also the same.

Both statements are true.

option C Correct 

Answer 5

• The product $AB$ is an (m×n)×(n×m)=m×m is a square matrix.
• The product $BA$ is an (n×m)×(m×n)=n×n is a square matrix.
• tr⁡(AB)=tr⁡(BA) is correct
• tr⁡(CD)=tr⁡(DC) is correct
• Both statements are Correct
   

Answer 6

Answer: C

$$\text{AB and BA share non-zero eigenvalues.}$$

$$\text{Non-zero eigenvalues of AB = Non-zero eigenvalues of BA}$$

$$\text{tr(AB) = Sum of eigenvalues of AB = Sum of non-zero eigenvalues of AB (0 eigen values do not change the sum)}$$

$$\text{= Sum of non-zero eigenvalues of BA = tr(BA)}$$

$$\text{Same is true for CD and DC.}$$

$$\text{Both statements are correct.}$$