1 1 vote $A$ is $5 \times 5$ matrix, all of whose entries are $1$ , then$A$ is not diagonalizable $A$ is idempotent $A$ is nilpotent $rank(A) = 1$ Linear Algebra goclasses goclasses-cs-dpp goclasses-cs-dpp-day-224 goclasses-da-dpp goclasses-da-dpp-day-126 linear-algebra goclasses-linear-algebra-practice-questions matrix + – GO Classes 410 views answer comment Share Follow Print See 1 comment 1 1 comment reply SivaKaliKeshav commented Apr 8 reply Follow flag $A$ is diagonalizable matix, note that it symmetric matrix $\implies \text{There exist an orthonormal basis of } \mathbb{R^5} \text{ such that it is a eigenbasis of A.}$ The eigen values of $A = \set{5, 0 , \ldots, 0}$ , Then $A = Q \begin{bmatrix}5 & 0 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \\ 0 & 0 &\ldots & 0 \\0 & 0 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \end{bmatrix}Q^T$ Then for any $k \in Z^+$, $A^k = Q\begin{bmatrix}5^k & 0 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \\ 0 & 0 &\ldots & 0 \\0 & 0 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \end{bmatrix}Q^T$ Since $5\neq0$, then $5^k \neq 0$, So for any positive $k$ matrix is $A^k$ is never a zero matrix. The only correct option is D. 0 0 replyShare Please log in or register to add a comment.
4 4 votes Option D pratishamacha answered Mar 21 pratishamacha comment Share Follow 0 reply Please log in or register to add a comment.
1 1 vote $A$ is symmetric (or selfadjoint, if your matrices are complex), so it is diagonalizable. It is not idempotent, because $A^2=5 A$. It is not nilpotent, because $A^n=5^{n-1} A$. All rows are identical and nonzero, so the row space is $1$-dimensional. Correct Answer : D GO Classes answered Mar 20 GO Classes comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes In a 5x5 matrix where every entry is 1, all the rows (and columns) are identical. Since there is only one unique, non-zero row, the rank of the matrix is 1. VIPIN_CHANDRA answered Mar 21 VIPIN_CHANDRA comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes A. Symmetric matrices are always diagonalizable. B. $A^2 = 5A$, not idempotent C. $A^k \neq 0$, not nilpotent D. $rank(A) = 1$, this is true. has only one linearly indepenent row. Answer: D Meticulous_March answered Mar 22 Meticulous_March comment Share Follow 0 reply Please log in or register to add a comment.
