0
$\begingroup$

I've got a matrix and linear mapping problem, giving me vector $v_1$ as \begin{bmatrix} 1\\ 0\\ 1\\ \end{bmatrix}

and $v_2$ as \begin{bmatrix} 0\\ 1\\ 1\\ \end{bmatrix} also, it gives me the mapping $f(x)=(x,v_1)v_1+(x,v_2)v_2$ while $x\in\mathbb{R}^3$ and the expression $(x,v_i)$ is dot product of $x$ and vector $v$

the problem asks me to find the basis of kernel of $f(x)$

so, I start thinking from $f(x)$. as dot product of vectors give scalar and I assume $x$ as \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} I'll get $f(x) = (x_1+x_3)\begin{bmatrix} 0\\ 1\\ 1\\ \end{bmatrix} + (x_2+x_3) \begin{bmatrix} 0\\ 1\\ 1\\ \end{bmatrix}$

and do the multiplication to get $f(x)=\begin{bmatrix} 1 & 0\\ 0 & 1\\ 1 & 1\\ \end{bmatrix}\begin{bmatrix} x_1+x_3\\ x_2+x_3\\ \end{bmatrix}$

and finally $f(x)=\begin{bmatrix} x_1+x_3\\ x_2+x_3\\ x_1+x_2+2x_3\\ \end{bmatrix}$

then to find kernel I should take the matrix equal to $0$ and find the kernel but I cannot find it

$\endgroup$

1 Answer 1

0
$\begingroup$

Guide:

$$f(x) = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2\end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}$$

Consider the matrix, $$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2\end{bmatrix}$$

Sum of the first two rows give us the third row and the first two rows are linearly independent.

Solve this linear system by letting $x_3=t$, express $x_1$ and $x_2$ in $t$. $$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}= \begin{bmatrix} 0 \\0 \\ 0\end{bmatrix}$$

$\endgroup$
1
  • $\begingroup$ I solved it and get the answer as $\begin{bmatrix} -1\\ -1\\ 1\\ \end{bmatrix}$. I understood your method well like I was really close. Thanks a lot $\endgroup$
    – RexSolus
    Aug 21, 2017 at 3:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.