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So the question asks:

find the kernel of the linear transformation $T : \mathbb{R}^4 \to \mathbb{R}^3$ defined by $T(x) = Ax$ where $A$ is the matrix: $$\begin{bmatrix}1 & 0 &1 & 0\\0 & 1 & 1 & 1\\1 & 1 & 1 & 1\end{bmatrix}$$

My solution is: $$\left[\begin{array}{cccc|c}1 & 0 &1 & 0 &0\\0 & 1 & 1 & 1 &0\\1 & 1 & 1 & 1 &0\end{array}\right]$$

$$ \begin{align} x_1+x_3 &= 0 \\ x_2+x_3+x_4 &= 0 \\ x_1+x_2+x_3+x_4 &= 0 \end{align} $$ which gives $x_1 = x_3 = 0$ and $x_2 + x_4 = 0$.

So my answer is kernel: $$\begin{bmatrix}0 \\t\\0\\-t\end{bmatrix}$$

Is it the right way to do this kind of question and is my answer correct?

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  • $\begingroup$ Yes, it is correct. $\endgroup$ – KittyL Feb 3 '16 at 11:39
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Note that from theorem of Nullity+Rank the dimension of $\Bbb R^4=4+dim \ker +dim Imf$ but the $\dim Imf=3$. Indeed the determinant of $$\begin{bmatrix}1 & 0 &1 \\0 & 1 & 1 \\1 & 1 & 1\end{bmatrix}$$ is $-1$. The $\ker $ is spanned by $(0,1,0-1)$

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