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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

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1 Answer 1

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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 $t$. $$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}= \begin{bmatrix} 0 \\0 \\ 0\end{bmatrix}$$

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  • $\begingroup$ 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

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