Finding kernel for a matrix I'm having trouble understanding what it means to find the kernel of a matrix.
I have a matrix as follows and I need to determine the kernel.
$\begin{bmatrix}2 & 4 & 5 \\ 1 & 2 & 3 \\ -1 & -1 & -3 \end{bmatrix}$
I'm able to get the augmented matrix down to echelon form as follows.
$\begin{bmatrix}1 & 2 & 3 & 0 \\ 0 & 5 & 4 & 0\end{bmatrix}$
Where do I take it from here? What would the form of the solution be?
Would I just write a couple of equations like the following two?
$x_1+2x_2+3x_3=0$
$0x_1+5x_2+4x_3=0$
 A: Finding the kernel of a matrix $A$ is finding the set of vectors that, when multiplied by A, result in the vector $0$. (It is easy to verify that this set of vectors is a vector space)
Mathematically speaking, you must solve the equation:
$Ax = 0$, where $x$ is an vector.
Note that this equation might have one solution or infinite solutions. In the second case, you'll find, while solving the system, a free variable, for example, $x_2$. Then, your other vectors can depend of this variable. You will find a free variable if you find something like $0 x_2 = 0$. Notice that you can have more than one free variable. The number of free variables is also the dimension of the null space of a transformation $T$ defined by $T(x) = Ax$.
Notice that, in your case, you have that your 3rd line is null. This means that $x_3$ is your free variable. Call it $t$ and write the solution in terms of it.
A: I found that this Khan Academy video was the best resource for helping me understand this problem more.
In the end I wrote the following for my answer.
$x_1=-\frac{7}{5}x_3+\frac{1}{5}x_4$
$x_2=-\frac{4}{5}x_3-\frac{3}{5}x_4$
I wasn't certain how to express this with $t$, as Dovah-king and the video mentioned, because we have two free variables, unlike the example in the video.
A: Let one of the variables, say $x_3$ be equal to $t$.
Then $5x_2 + 4t = 0, x_2 = -\frac{4}{5}t$ and $x_1 + 2x_2 + 3x_3 = 0$ or $x_1 = -3t - 2 \cdot -\frac{4}{5}t = -\frac{7}{5}t$.
This means that $\begin{pmatrix} x_1 \cr x_2 \cr x_3 \end{pmatrix} = \begin{pmatrix} -7/5 \cr -4/5 \cr 1 \end{pmatrix} t$. We have parameterised all the variables in terms of a single parameter, which means the kernel has dimension $1$.
As a quick check using the rank-nullity theorem, $\text{dim} (\text{ker} (A)) + \text{rank}(A) = n$, which is indeed true as $1 + 2 = 3$.
