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I am asked to find the basis of the kernel of $\beta$, where $\beta$: $\mathbb{R^3}$ $\rightarrow$ $\mathbb{R^3}$,

$$ \beta (x,y,z) = \begin{pmatrix} x+y \\ 2x+y-z \\ x-z \\ \end{pmatrix} $$

I've put the matrix $$ \beta (x,y,z) = \begin{pmatrix} 1&1&0 \\ 2&1&-1\\ 1&0&-1 \\ \end{pmatrix} $$ into row reduced echelon form and ended up with the standard 3x3 identity matrix. Is the basis therefore

$$ \beta (x,y,z) = \begin{pmatrix} 1 \\ 1\\ 1 \\ \end{pmatrix} $$

or have I done this incorrectly? I would say this is a simple example but the answer I have is different to the one given which is $(1,-1,1)$. Thanks in advance.

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First of all, if you indeed end up with identity matrix, it means that $\beta$ is of full rank, i.e. isomorphism, which would mean that the kernel is trivial. This makes your conclusion invalid. What you have to do is solve linear equation $\beta x = 0$. When you reduce the matrix you finally get \begin{pmatrix} 1 & 0 & -1\\ 0& 1 & 1\\ 0 & 0 & 0 \end{pmatrix}

So, the matrix is actually of rank $2$, which means that kernel is one-dimensional. It is spanned by $(1,-1,1)$ because, as you can check, it indeed solves the system with above matrix. What we did is look at last column, multiplied it by $-1$ and replaced $0$ with $1$ (standard procedure when solving linear systems).

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Row reducing the matrix should've given you the matrix $$ \begin{pmatrix} 1&1&0 \\ 0&1&1 \\ 0&0&0\\ \end{pmatrix} $$ rather than the identity matrix, since the determinant of the matrix is zero: the second row is the sum of the first and third rows!

It is then easy to see that a vector $(x,y,z)$ is an element of the kernel if and only if $x+y = 0$ and $y+z = 0$.

From these equations, you can derive that $$\operatorname{ker}(\beta) = \{(x,-x,x) \mid x \in \mathbb{R}\},$$ which is generated by the vector $(1,-1,1)$ as per your solutions.

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