Answer to Question #101035 in Linear Algebra for Francis

Question

1. For purposes of this problem, let us define a "checkerboard matrix" to be a square matrix A = [aij] such that

a Subscript i j Baseline equals Start Layout left-brace 1st Row 1st Column 1 if i+j is even 2nd Row 1st Column 0 if i+j is odd End Layout

Find the rank and nullity of the following checkerboard matrix.

The 6 × 6 checkerboard matrix.

a. Rank (A) =
b. nullity (A) =

2. For purposes of this problem, let us define a "checkerboard matrix" to be a square matrix Upper A equals left-bracket a Subscript i j Baseline right-bracket such that

a Subscript i j Baseline equals Star tLayout left-brace 1st Row 1st Column -2 if i plus j is even 2nd Row 1st Column 0 if i plus j is odd End Layout

Find the rank and nullity of the n times n checkerboard matrix for n greater-than-or-equal-to 2.

a. Rank (A) =
b. nullity (A) =

Answer 1

1. $A=\begin{bmatrix} 1&0&1&0&1&0\\ 0&1&0&1&0&1\\ 1&0&1&0&1&0\\ 0&1&0&1&0&1\\ 1&0&1&0&1&0\\ 0&1&0&1&0&1 \end{bmatrix}$ is the 6x6 checkerboards matrix.

Applying $R_6 \gets R_6-R_2$ ; $R_4 \gets R_4-R_2$ ; $R_5 \gets R_5-R_1$ and $R_3 \gets R_3-R_1$ ; we get;

$A=\begin{bmatrix} 1&0&1&0&1&0\\ 0&1&0&1&0&1\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{bmatrix}$ . Clearly, rank(A)=2.

Nullity is given by; Ax=0; where x represents Null Space of A.

$\begin{bmatrix} 1&0&1&0&1&0\\ 0&1&0&1&0&1\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{bmatrix}\begin{bmatrix} x_1 \\x_2\\x_3\\x_4\\x_5\\x_6 \end{bmatrix} =\begin{bmatrix} 0 \\0\\0\\0\\0\\0 \end{bmatrix}$

$\implies x_1+x_3+x_5=0;\implies x_1=-x_3-x_5;$

$\implies x_2+x_4+x_6=0;\implies x_2=-x_4-x_6$

$\begin{bmatrix} x_1 \\x_2\\x_3\\x_4\\x_5\\x_6 \end{bmatrix}=\begin{bmatrix} -x_3-x_5\\-x_4-x_6\\x_3\\x_4\\x_5\\x_6 \end{bmatrix}$ . Thus, 4 independent parameters, $x_3,x_4,x_5,x_6\implies Nullity(A)=4$

2.$A=\begin{bmatrix} -2&0&-2&0&-2&0\\ 0&-2&0&-2&0&-2\\ -2&0&-2&0&-2&0\\ 0&-2&0&-2&0&-2\\ -2&0&-2&0&-2&0\\ 0&-2&0&-2&0&-2 \end{bmatrix}$ is the checkerboard matrix (for n=6).

Applying $R_6 \gets R_6-R_2; R_4 \gets R_4-R_2; R_5 \gets R_5-R_1and R_3 \gets R_3-R_1$ we get;

$A=\begin{bmatrix} -2&0&-2&0&-2&0\\ 0&-2&0&-2&0&-2\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{bmatrix}$ Clearly, rank(A)=2. Moreover this applies to all matrices A for n>2. As we can perform the same row transformations again and again.

$\implies Rank(A)=2$

Similarly;$Nullity(A)=4$