# Full rank vs short rank matrix

I am given the definition: "A matrix A is of full rank if and only if the vector $d$ for which $Ad=0$ is $d=0$."

I don't understand: if we have the matrix
$$\begin{pmatrix}1&2&3\\ 4&5&6\\ 13&19&88\end{pmatrix}$$
It is not of full rank, but what number other than $0$ can we multiply it by to get $0$? The last line is just an example that is independent of the first two.

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The only vector $d$, not number. – Cameron Buie Oct 2 '12 at 17:06
Full rank means that the columns of the matrix are independent; i.e., no column can be written as a combination of the others. – saadtaame Oct 2 '12 at 17:07

You don't multiply by a number, but by a vector (on the right). The matrix in your example in fact is of full rank, so I can't give an example there, but if we instead take the matrix:

$$\begin{pmatrix}1&2&13\\2&4&19\\3&6&88\end{pmatrix}$$

which is not of full rank, then multiplying on the right by $(-2,1,0)$ gives:

$$\begin{pmatrix}1&2&13\\2&4&19\\3&6&88\end{pmatrix}\begin{pmatrix}-2\\1\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$

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Thank you for the helpful example. – metrix Oct 2 '12 at 17:15

Full rank means that the columns of the matrix are independent; i.e., no column can be written as a combination of the others. When you multiply a matrix by a vector (right), you are actually taking a combination of the columns, if you can find at least one vector such that the multiplication gives the 0 vector, then the columns are dependent and the matrix is not full rank.

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Thank you. Very useful. – metrix Oct 2 '12 at 17:16