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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?

  1. Statement $1$ is correct and Statement $2$ is wrong.
  2. Statement $1$ is wrong and Statement $2$ is correct.
  3. Both Statement $1$ and Statement $2$ are correct.
  4. Both Statement $1$ and Statement $2$ are wrong.
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4 Answers

Best answer
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35 votes

Answer C

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-

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Statement -1 is evident from the definition of matrix product 

{\displaystyle \operatorname {tr} (\mathbf {A} \mathbf {B} )=\sum _{i=1}^{m}\left(\mathbf {A} \mathbf {B} \right)_{ii}=\sum _{i=1}^{m}\sum _{j=1}^{n}a_{ij}b_{ji}=\sum _{j=1}^{n}\sum _{i=1}^{m}b_{ji}a_{ij}=\sum _{j=1}^{n}\left(\mathbf {B} \mathbf {A} \right)_{jj}=\operatorname {tr} (\mathbf {B} \mathbf {A} ).}

 

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

 

So, both are TRUE 

option C

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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:

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