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Find a linear transformation $T : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ such that

$$ \operatorname{Ker} (T) = [(1, 0, 0, 1) ; (-1, 0, 0, 1)] $$

$$ \text{Image}(T) = \operatorname{Im}(T) = [(1, -1, 0, 2); (0, 1, -1, 0)] $$

Thoughts on this problem:

Given a linear transformation $T$, I know how to find its Kernel/null space and its image. However, I don't know how to "reverse engineer" a linear transformation such that I will obtain these fixed Kernel and Image. So I don't know how to proceed here and have no work to show.

Thanks.

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You know that the linear transformation

$$ T: \mathbb{R}^4 \to \mathbb{R}^4: \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} \mapsto T \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} $$

has vectors $Ker(T)=[(1,0,0,1);(−1,0,0,1)]$ as kernel.

The span of these two vectors is equal to $$\begin{bmatrix} \alpha \\ 0 \\ 0 \\ \beta \end{bmatrix}$$

so the first and fourth vectors of the standard basis get mapped to zero.

Now, for the matrix of a linear transformation with respect to the standard basis, the first column is the image of the first basis vector, the second column is the image of the second basis vector and so on. Therefore, the first and fourth column of your $T$ must be zero, as the first and fourth basis vector lie in the kernel.

Expanding on this idea, it now suffices to choose the second column and third column as the given two vectors in the image $Im(T)=[(1,−1,0,2);(0,1,−1,0)]$, as this ensure that there is a domain vector which gets mapped to them.

So in conclusion, $T$ is (non-uniquely!)

$$ T = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 2 & 0 & 0 \end{bmatrix}.$$

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Let $T$ be associated with some matrix $A$. The first constraint fixes $$ A((1,0,0,1)^T) = (0,0,0,0)^T = A((-1,0,0,1)^T) $$

The second will work as long as you pick any two vectors $\vec{v},\vec{w}$, which together with $\Ker(T)$ will be a linearly independent set, and force $A\vec{v} = (1,-1,0,2)^T$ and $A\vec{w} = (0,1,-1,0)^T$.

Solve these four equations to get your $A$...

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