Show that linear transformation is not surjective? Given the matrix, $$M =
\begin{bmatrix}1&7&9&3\\2&15&19&8\\7&52&66&27\\3&4&10&-24\end{bmatrix}$$
Show that the linear transformation $T_m: \mathbb R^4 \to \mathbb R^4$ defined by the multiplication of column vectors of the left by $M$ is not surjective by exhibiting a column vector 
\begin{bmatrix}a\\b\\c\\d\end{bmatrix}
not in the image of $T_m$, ie such that
$$M*\begin{bmatrix}x\\y\\z\\t\end{bmatrix}=
\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$
Info needed:
row reduced M= \begin{bmatrix}1&0&2&0\\0&1&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}
row reduced $M^T$ \begin{bmatrix}1&0&1&0\\0&1&3&0\\0&0&0&1\\0&0&0&0\end{bmatrix}

A very important identity that you may have seen in your course so far is  $$\text{Im}(M)^{\perp}=\text{Ker}\left(M^{T}\right).$$
 This is true because if $\vec{y}\in\text{Im}(A)^\perp$ then by definition it
  satisfies $\langle\vec{y}\cdot M\vec{x}\rangle=0$
  for all $\vec{x}$ and so $$\langle\vec{y}\cdot M\vec{x}\rangle=\vec{y}^{T}M\vec{x}=\left(M^{T}\vec{y}\right)^{T}\vec{x}=\langle M^{T}\vec{y},\vec{x}\rangle=0$$
 for all $\vec{x}$
 , and so $M^{T}\vec{y}$ must be the all zero vector, and hence $\vec{y}\in\text{Ker}(M^{T})$. Now, since $\text{Im}(M)^{\perp}=\text{Ker}\left(M^{T}\right)$, any nontrivial vector in $\text{Ker}(M^{T})$
  cannot be in the image of $M$
  since it lies in the orthogonal complement of the image. So you need only find a non-zero vector in $\text{Ker}(M^{T})$.
