Answer to Question #101033 in Linear Algebra for Francis Joco

Question

1. 1 3 6 0
A = -2 4 6 1
-3 1 0 1

(a) Find an equation relating nullity ⁢ left-parenthesis Upper A right-parenthesis and ⁢ nullity left-parenthesis Upper A Superscript Upper T Baseline right-parenthesis for the matrix A.

(b) Find an equation relating nullity ⁢ left-parenthesis Upper A right-parenthesis and ⁢ nullity left-parenthesis Upper A Superscript Upper T Baseline right-parenthesis for a general m times n matrix.

2. Are there values of r and s for which

Close and open Parenthesis 1 0 0
0 r-6 6
0 s-5 r+6
0 0 7

has rank 1 or 2? If so, find those values.
The matrix has ____ for

r = ?
s = ?

Answer 1

1)

a) $A=\begin{pmatrix} 1&3&6&0 \\ -2&4&6&1 \\ -3&1&0&1 \end{pmatrix}, \; x=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}$

$Ax=0$

$\begin{cases} x_1+3x_2+6x_3+0=0 \\ -2x_1+4x_2+6x_3+x_4=0 \\ -3x_1+x_2+0+x_4=0 \end{cases} \begin{cases}x_1+3x_2+6x_3+0=0 \\ 0+10x_2+18x_3+x_4=0 \\ 0+10x_2+18x_3+x_4=0 \end{cases}$

$\begin{cases} x_1=-3x_2-6x_3=-3/5x_3+3/10x_4 \\ x_2=-9/5x_3-1/10x_4 \end{cases}$

$x=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}=\begin{pmatrix} -3/5x_3+3/10x_4 \\ -9/5x_3-1/10x_4 \\ x_3 \\ x_4 \end{pmatrix}= x_3 \begin{pmatrix} -3/5 \\ -9/5 \\ 1 \\0 \end{pmatrix}+ x_4 \begin{pmatrix} 3/10 \\ -1/10 \\ 0 \\1 \end{pmatrix} \Rightarrow$

$\Rightarrow$ nullity(A)=2 (because nullity (A) = the number of parameters in the solution of Ax = 0)

$A^T=\begin{pmatrix} 1&-2&-3 \\ 3&4&1 \\ 6&6&0 \\ 0&1&1 \end{pmatrix} , x=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$

$A^Tx=0$

$\begin{cases} x_1-2x_2-3x_3=0 \\ 3x_1+4x_2+x_3=0 \\ 6x_1+6x_2+0=0 \\ 0+x_2+x_3=0 \end{cases} \begin{cases} x_1-2x_2-3x_3=0 \\ 0+10x_2+10x_3=0 \\ 0+18x_2+18x_3=0 \\ 0+x_2+x_3=0 \end{cases} \begin{cases} x_1-2x_2-3x_3=0 \\ 0+x_2+x_3=0 \end{cases}$

$\begin{cases}x_1=2x_3+3x_3=x_3 \\ x_2=-x_3\end{cases}$

$x=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} x_3 \\- x_3 \\ x_3 \end{pmatrix} = x_3 \begin{pmatrix}1 \\ -1\\ 1 \end{pmatrix} \Rightarrow$

$\Rightarrow$ nullity(A^T)=1 (because nullity (A) = the number of parameters in the solution of Ax = 0)

nullity(A)+nullity(A^T)=3 (=3+4-2*2)

b) We can use formula:

• for any mxn matrix A: rank (A) + nullity (A) = n

• for any matrix A: rank (A) = rank (A^T)

rank(A)+nullity(A)=n

rank(A^T)+nullity(A^T)=m

rank(A)+rank(A^T)+nullity(A)+nullity(A^T)=m+n

nullity(A)+nullity(A^T)=m+n-2*k, where k=rank(A)=rank(A^T)

2) $B=\begin{pmatrix}1&0&0 \\ 0&r-6&6 \\ 0&s-5&r+6 \\ 0&0&7\end{pmatrix}$

rank (B) = dimension of its row space (or column space)

We can see that 1 and 4 rows are linear independent $\Rightarrow$ rank (B) $\geq$ 2

So, rank (B) can’t be equal to 1, there are no values of r and s for which rank(B)=1

Suppose that rank(B)=2.

1 and 4 rows are linear independent, rank(B)=2 $\Rightarrow$ they are basis vectors of row space $\Rightarrow$

$\Rightarrow \exists a,b: \begin{pmatrix} 0\\r-6\\6\end{pmatrix} =a \begin{pmatrix} 1\\0\\0\end{pmatrix}+b \begin{pmatrix} 0\\0\\7\end{pmatrix},$

$\begin{cases} a=0 \\ 0=r-6 \\7b=6 \end{cases} \begin{cases} a=0 \\b=6/7\\r=6 \end{cases}$

$\Rightarrow \exists a,b: \begin{pmatrix} 0\\s-5\\r+6\end{pmatrix} =a \begin{pmatrix} 1\\0\\0\end{pmatrix}+b \begin{pmatrix} 0\\0\\7\end{pmatrix},$

$\begin{cases} a=0 \\ 0=s-5\\7b=r+6=12 \end{cases} \begin{cases} a=0 \\b=12/7\\s=5 \end{cases}$

$B=\begin{pmatrix}1&0&0 \\ 0&0&6 \\ 0&0&12\\ 0&0&7\end{pmatrix}$ has rank=2 (for r=6,s=5)