Question-71

\(\displaystyle A\) is a square matrix. Consider the two statements given below:

\(\displaystyle P\): \(\displaystyle A\) and \(\displaystyle A^{T}\) have the same eigenvalues.

\(\displaystyle Q\): \(\displaystyle A\) and \(\displaystyle A^{T}\) have the same eigenvectors.

Which of the following is true?

• \(\displaystyle P\) is true, but \(\displaystyle Q\) is false.

• \(\displaystyle P\) is false, but \(\displaystyle Q\) is true.

• Both \(\displaystyle P\) and \(\displaystyle Q\) are true.

• Both \(\displaystyle P\) and \(\displaystyle Q\) are false

Answer

• \(\displaystyle P\) is true, but \(\displaystyle Q\) is false.

• \(\displaystyle P\) is false, but \(\displaystyle Q\) is true.

• Both \(\displaystyle P\) and \(\displaystyle Q\) are true.

• Both \(\displaystyle P\) and \(\displaystyle Q\) are false

Solution

Let us consider the characteristic polynomial of \(\displaystyle A^{T}\):

\[ \begin{equation*} \begin{aligned} |A^{T} -\lambda I| & =|A^{T} -\lambda I^{T} |\\ & =|( A-\lambda I)^{T} |\\ & =|A-\lambda I| \end{aligned} \end{equation*} \]

We see that both \(\displaystyle A\) and \(\displaystyle A^{T}\) have the same characteristic polynomial. Therefore, they have the same eigenvalues.

To see that they don’t need to have the same eigenvectors, consider the following example:

\[ \begin{equation*} A=\begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} ,\ A^{T} =\begin{bmatrix} 1 & 0\\ 1 & 2 \end{bmatrix} \end{equation*} \]

The eigenvalues of \(\displaystyle A\) and \(\displaystyle A^{T}\) are \(\displaystyle 1,2\). A pair of eigenvectors of \(\displaystyle A\) is \(\displaystyle ( 1,0)\) and \(\displaystyle ( 1,1)\). A pair of eigenvectors for \(\displaystyle A^{T}\) is \(\displaystyle ( 1,-1)\) and \(\displaystyle ( 0,1)\).