Find the kernel of a 4x4 matrix

Question

$$ \begin{pmatrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12\\ 13 & 14 & 15 & 16\\ \end{pmatrix} $$

I am asked to find the kernel of the matrix $M$. After doing some row operation I get to $$ \begin{pmatrix} 1 & 2 & 3 & 4\\ 0 & -4 & -8 & -12\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} $$

and for $x$ I find $x = \alpha + 2\beta$, whereas $y = -2\alpha -3\beta$ Therefore, $$ \begin{pmatrix} \alpha + 2\beta\\ -2\alpha - 3\beta\\ \alpha\\ \beta\\ \end{pmatrix} $$

When we take outside alpha and beta: we get two vectors: $$ \begin{pmatrix} 1\\ -2\\ 1\\ 0\\ \end{pmatrix} $$ and $$ \begin{pmatrix} 2\\ -3\\ 0\\ 1\\ \end{pmatrix} $$ which are linearly independent and form a basis of this $ker(M)$ Could you please confirm with me whether you get the same result? Thank you.

Answer 1

Yes it is correct, in case of doubt you can check it directly by simple multiplication for RREF matrix

$$\begin{bmatrix} 1 & 2 & 3 & 4\\ 0 & -4 & -8 & -12\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}\begin{bmatrix} 1\\ -2\\ 1\\ 0\\ \end{bmatrix}=0$$

$$\begin{bmatrix} 1 & 2 & 3 & 4\\ 0 & -4 & -8 & -12\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}\begin{bmatrix} 2\\ -3\\ 0\\ 1\\ \end{bmatrix} =0 $$

or/and also for the original matrix

$$\begin{bmatrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12\\ 13 & 14 & 15 & 16\\ \end{bmatrix} \begin{bmatrix} 1\\ -2\\ 1\\ 0\\ \end{bmatrix}=0$$

$$\begin{bmatrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12\\ 13 & 14 & 15 & 16\\ \end{bmatrix} \begin{bmatrix} 2\\ -3\\ 0\\ 1\\ \end{bmatrix} =0 $$

Note that also after this check, for the correctness of the result it is crucial that RREF is calculated properly. In case of doubt on it we can evaluate $rank(M)$ with others methods or follow the nice suggestion give by Atmos to be sure that $dim(Ker M)=2$ and thus that what we have found is a basis for $Ker(M)$.

Answer 2

Your computations are correct and so is your answer. But you should really improve your MathJax skills.

Answer 3

Yes it looks good great. However, you should precise why the dimension of Kernel is $2$. For example by extracting the $2 \times 2$ determinant $$ \begin{vmatrix} 3&4 \\ 7&8 \end{vmatrix}=24-28 \ne 0 $$ So the dimesion of the image is at least $2$. Then you found two vectors in it, so the dimension of the image cannot exceed two so it values $2$. as well as the dimension of the Kernel.