Conditions that ensure Null(A) = {0}

Question

Just would like some clarification for this question:

Let A and B be the matrices

A =

\begin{bmatrix} 1 & 1 & \alpha \\ 1 & 0 & 0 \\ 0 & 2 & \beta \end{bmatrix} and

B= \begin{bmatrix} 1 & 1 & \beta \\ \alpha & 0 & 0 \\ 0 & 3 & \alpha \end{bmatrix}

(a) Find conditions on $\alpha$ and $\beta$ that ensures Null(A) = {0}

(b) Find conditions on $\alpha$ and $\beta$ that ensures Col(B) = $\mathbb{R^3}$

So for (a) so far i reduced the matrix up to:

$\overset{R_2=R_2-R_1}{\longrightarrow}$

$\overset{R_2=-R_2}{\longrightarrow}$

$\overset{R_3=R_3-2R_2}{\longrightarrow}$

\begin{bmatrix} 1 & 1 & \alpha \\ 0 & 1 & \alpha \\ 0 & 0 & \beta-2\alpha \end{bmatrix}

So would the conditions be $\alpha \neq -1$, thus $\beta \neq -2$ so it equals a zero vector??

and for (b) i did:

where $a\begin{bmatrix} 1 \\ \alpha \\ 0 \end{bmatrix} +b\begin{bmatrix} 1\\ 0 \\ 3 \end{bmatrix}+c\begin{bmatrix}\beta\\ 0\\ \alpha\end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$

\begin{bmatrix} 1 & 1 & \beta |x\\ \alpha & 0 & 0 |y\\ 0 & 3 & \alpha |z \end{bmatrix}

$\overset{R_2= (1/\alpha) R_2}{\longrightarrow}$

$\overset{R_2=R_2-R_1}{\longrightarrow}$

$\overset{R_2=-R_2}{\longrightarrow}$

$\overset{R_3=R_3-3R_2}{\longrightarrow}$

= \begin{bmatrix} 1 & 1 & \beta |x\\ 0 & 1 & \beta |x-y \\ 0 & 0 & \alpha-3\beta | z-3(x-y) \end{bmatrix}

I think overall i'm just abit confused about the concept of Column Spaces :/ Would appreciate any help thankyou!

Answer 1

If your RREF is $$\begin{bmatrix} 1 & 1 & \alpha \\ 0 & 1 & \alpha \\ 0 & 0 & \beta-2\alpha \end{bmatrix}$$

to ensure that the matrix is nonsingular, we just require $\beta-2\alpha \ne 0$. Just check that by imposing this condition, the determinant is non-zero.

For the second equation, for the matrix to span $\mathbb{R}^3$, it is essentially requiring the matrix to have the kernel to be $\{0\}$ again.

In one of the elementary operation, we divide by $\alpha$, hence, we should check what happens if $\alpha=0$, if $\alpha=0$, then we will have a zero row and the matrix can't be of full rank.

Hence, we need to impose the condition that $\alpha \ne 0$, looking at the RREF, we also require $\alpha - 3\beta \ne 0$.

Answer 2

$\text{Null}(A) = \{0\} \Leftrightarrow \det(A) = 2\alpha - \beta \neq 0$.

$B \in \mathbb{R}^{3\times 3}$ and $\text{Col}(B) = \mathbb{R}^{3} \Leftrightarrow \det(B) = \alpha(3\beta - \alpha) \neq 0$.