Answer to Question #101036 in Linear Algebra for Santos

Question

1. For purposes of this exercise, let us define an " X -matrix" to be a square matrix with an odd number of rows and columns that has 0's everywhere except on the two diagonals, where it has 1's. Find the rank and nullity of the following X -matrix.

close and open parenthesis 1 0 1
0 1 0
1 0 1

rank (X) =
nullity (X) =

Answer 1

$Let A=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 &0\\ 1&0&1 \end{bmatrix}$ be the given matrix. Performing the operation $R_3 \gets R_3-R_1$ we get;

$A=\begin{bmatrix} 1 & 0&1 \\ 0 & 1&0\\ 0&0&0 \end{bmatrix}$ Clearly, the largest sub matrix with non-zero determinant is of order $2\Chi2$ .

$\implies rank(A)=2$ (Answer)

Null space of a matrix A is x given by the equation $Ax=0$

$\begin{bmatrix} 1 & 0&1 \\ 0 & 1&0\\ 0&0&0 \end{bmatrix}.\begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}$ $\implies \begin{bmatrix} x_1+x_3 \\x_2\\0 \end{bmatrix} =\begin{bmatrix} 0\\0\\0\end{bmatrix}$ -----(i)

Thus, null space(A) is given by; $x=\begin{Bmatrix} x_1 & x_2&x_3 \end{Bmatrix}$ where $x_1,x_2,x_3$ follow(using (i)):

$x_1+x_3=0\implies x_3=-x_1$ -----(ii)

$x_2=0$ ------(iii)

$\implies nullspace(A)=\begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix}x_1\\0\\-x_1 \end{bmatrix}=$$x_1\begin{bmatrix} 1\\0\\-1 \end{bmatrix}=span(\begin{Bmatrix} 1,0,-1\end{Bmatrix})$ (from (ii) and (iii))

Thus, nullity(A)=dimension of null space(A)

=1 (Answer)