# Is this still considered $\{0\}$

Is the following solution to the matrix a zero subspace? (Assume that the last column of zeros is the constant portion of the matrix)

I'm working on some kernel problems, and if a linear transformation is one-to-one or not, I have to see if the kernel of the transformation is the zero subspace or not.

A=$\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$

(If anything I said is unclear, assume that the first column is $x$ and second column is $y$ and third column is for constants. The first row says $x=0$, the second row says $y=0$)

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What does the third row represent? An augmented matrix with two variables would have two rows and three columns, not three rows and three columns. – J Swanson Jul 30 '13 at 21:45

This matrix is of rank $2$, which by the fundamental theorem of linear algebra means that the kernel is a $3-2=1$ dimensional subspace, so it cannot be the 0 vector.
$$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$
You are correct that $x=0$ and $y=0$, but what about $z$?