# Find a 3x3 Matrix given its Kernel, an element in its Image and one of its Eigenvectors

The question asks to find a 3x3 matrix, knowing:

1. Its Kernel is the span of $$\begin{bmatrix}{1\\1\\1}\end{bmatrix}$$
2. The vector $$\begin{bmatrix}{1\\2\\-1}\end{bmatrix}$$ is in its Image (column space)
3. $$\begin{bmatrix}{1\\0\\1}\end{bmatrix}$$ is one of its eigenvectors

I don't know the definite way to do problems like this, so how I did it requires a bit of guessing. From everything I know so far, here's my train of thoughts:

So, first of all, the Kernel is the span of one 3D vector aka. a 3D line through the origin, the Kernel has the dimension of 1, the rank-nullity theorem suggests the Image or column space of that matrix has dimension 2. Thus the column space is a plane going through the origin, and the vector $$\begin{bmatrix}{1\\2\\-1}\end{bmatrix}$$ is on the plane. Also, the matrix is not in full rank, because its rank is 2. This implies its determinant must be zero, so 0 is one of its eigenvalues. We can imply from the first clue that $$\begin{bmatrix}{1\\1\\1}\end{bmatrix}$$ is also an eigenvector corresponding to the eigenvalue 0, together with $$\begin{bmatrix}{1\\0\\1}\end{bmatrix}$$ (but I'm not sure if these two vectors are sufficient to form an eigenbasis, later on it turns out to be not).

Suppose $$\begin{bmatrix}{1\\0\\1}\end{bmatrix}$$ has a non zero corresponding eigenvalue. Because a matrix only scales its eigenvector if the eigenvalue is not zero, $$\begin{bmatrix}{1\\0\\1}\end{bmatrix}$$ is in its Image plane. Choose $$\begin{bmatrix}{1\\0\\1}\end{bmatrix}$$ and $$\begin{bmatrix}{1\\2\\-1}\end{bmatrix}$$, which are linearly independent, as the basis of that Image plane, put that into our matrix, we got two of its column. The third must be some linear combination of the first two.

Here's when the guessing comes. First, I guess it's this matrix: $$\begin{bmatrix}1 & 1 & -2\\0 & 2 & -2\\1 & -1 & 0\end{bmatrix}$$

It has the right column space and kernel. But the eigenvectors are found to be: $$\begin{bmatrix}{1\\1\\1}\end{bmatrix}$$, $$\begin{bmatrix}{3\\2\\1}\end{bmatrix}$$ and $$\begin{bmatrix}{1\\1\\0}\end{bmatrix}$$.

So I switched the first two columns: $$\begin{bmatrix}1 & 1 & -2\\2 & 0 & -2\\-1 & 1 & 0\end{bmatrix}$$

This time the matrix' eigenvectors are correct: $$\begin{bmatrix}{1\\1\\1}\end{bmatrix}$$, $$\begin{bmatrix}{1\\0\\1}\end{bmatrix}$$ and $$\begin{bmatrix}{1\\1\\0}\end{bmatrix}$$, also its kernel and column space satisfy the problem. We can say this is one answer.

However, I'm pondering: Is there any method to do this without guessing? And is that matrix the only answer to this question?

That said, here's a way to do things systematically. If we select a matrix so that its eigenvectors are $$(1,1,1),(1,2,-1),(1,0,1)$$ with associated eigenvalues $$0,\lambda_1,\lambda_2$$ (where $$\lambda_1$$ and $$\lambda_2$$ are both non-zero), then the resulting matrix will have the required properties.
If we take $$D = \pmatrix{0 \\ & \lambda_1 \\ && \lambda_2}, \quad S = \pmatrix{1&1&1\\1&2&0\\1&-1&1},$$ then we can take the matrix $$A = SDS^{-1}$$. If you want to try this for yourself, the inverse of $$S$$ is given by $$S^{-1} = \frac 12 \pmatrix{-2 & 2 & 2\\ 1 & 0 & -1\\ 3 & -2 & -1}.$$ With the help of block-matrix multiplication, we can also take advantage of the zero in the diagonal matrix $$D$$ to write this as $$A = SDS^{-1} = \frac 12 \pmatrix{1&1\\2&0\\-1&1} \pmatrix{\lambda_1\\ & \lambda_2} \pmatrix{1&0&-1\\3&-2&-1}.$$ For instance, taking $$\lambda_1 = \lambda_2 = 1$$ yields the answer $$A = \pmatrix{2 & -1 & -1\\ 1 & 0 & -1\\ 1 & -1 & 0}.$$