GATE Data Science and Artificial Intelligence 2024 | Sample Paper | Question: 48

Question

Consider the matrix $\mathbf{X}$ whose eigenvalues are $1,-1$ and $3$. Then Trace of $\mathbf{X}^{3}-3 \mathbf{X}^{2}$ is __________.

Comments

$Ax = \lambda x \implies A^nx = \lambda ^n x$

$\implies A^2x = AAx = A \lambda x = \lambda Ax = \lambda \lambda x = \lambda ^2x$

Answer 1

Ans:

Answer 2

by cayley hamilton theorem we know that the matrix also satisfies the characterstic equation

$$(\lambda – 1)(\lambda + 1)(\lambda – 3)=0 \longrightarrow (X – 1)(X + 1)(X – 3)=0$$

we get $$X^3 -3X^2 – X  + 3I = 0 \longrightarrow X^3 -3X^2 = X  - 3I$$

so lets first find the dimensions of $X$,as X has 3 eigen values it is a 3x3 matrix

$$tr(X-3I) = tr(X)  -3tr(I)$$ where $trace(X) = \text{sum of its eigen values}$ and $I$ has eigen values $1,1,1$

so $ans = tr(X-3I) = (1-1+3) \ – \  3(3) = -6$

Answer 3

Let \(X\) be the matrix whose eigenvalues are \(1, -1,\) and \(3\). Denote these eigenvalues as \(λ_1 = 1\), \(λ_2 = -1\), and \(λ_3 = 3\). The expression \(X^3 - 3X^2\) for a diagonalizable matrix with eigenvalues \(λ_i\) is given by: \[ X^3 - 3X^2 = P D^3 P^{-1} - 3 P D^2 P^{-1} \] Here, \(P\) is the matrix of eigenvectors and \(D\) is the diagonal matrix of eigenvalues. Substituting the given eigenvalues: \[ X^3 - 3X^2 = P \begin{bmatrix} 1^3 & 0 & 0 \\ 0 & (-1)^3 & 0 \\ 0 & 0 & 3^3 \end{bmatrix} P^{-1} - 3 P \begin{bmatrix} 1^2 & 0 & 0 \\ 0 & (-1)^2 & 0 \\ 0 & 0 & 3^2 \end{bmatrix} P^{-1} \] Simplifying further: \[ X^3 - 3X^2 = P \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 27 \end{bmatrix} P^{-1} - 3 P \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{bmatrix} P^{-1} \] Now, evaluating the trace: \[ \text{Trace}(X^3 - 3X^2) = \text{Trace}(P \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 27 \end{bmatrix} P^{-1}) - 3 \text{Trace}(P \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{bmatrix} P^{-1}) \] Simplifying further: \[ \text{Trace}(X^3 - 3X^2) = \text{Trace}\left(\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 27 \end{bmatrix}\right) - 3 \text{Trace}\left(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{bmatrix}\right) \] Finally: \[ \text{Trace}(X^3 - 3X^2) = (1 - 1 + 27) - 3(1 + 1 + 9) = 27 - 3(11) = -6 \] Therefore, the trace of \(X^3 - 3X^2\) is \(-6\).