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I was trying to do my Linear algebra HW, and I was stuck on a question. The question is the one shown below: Linear Algebra

I was able to do part (a), but I am having trouble in doing part (b) and (c). For part (b), I assumed an arbitrary matrix $ \ A= \left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] $, and applied the linear transformation F(A), which gave me $ \ F(A)= \left[ {\begin{array}{cc} -a+2 b \\ -c+2d \\ \end{array} } \right]$. In order for it be a kernel, it must equal the zero vector, thus a=2b, c=2d. Thus $ \ A= \left[ {\begin{array}{cc} 2b & b \\ 2d & d \\ \end{array} } \right]$

But does that mean that the ker F= {$ \ \left[ {\begin{array}{cc} 2 \\ 0 \\ \end{array} } \right]$, $ \ \left[ {\begin{array}{cc} 0 \\ 1 \\ \end{array} } \right]$} as these will span the whole of $R^2$ and this is what I get after separating b and d vectors. Am I doing something wrong here.

Also, how to find a coordinate vector [B] with respect to the basis that I got (or will get if I am wrong). Any help will be greatly appreciated.

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  • $\begingroup$ The kernel of a linear map $f: X \to Y$ belongs to $X$, not $Y$. $\endgroup$ – nicomezi Jan 29 '18 at 17:43
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    $\begingroup$ so, does that mean ker F should span whole of $R^(2*2)$ not just $R^2$ $\endgroup$ – Tumul Kumar Jan 29 '18 at 17:45
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    $\begingroup$ The basis of the kernel must span the kernel, not necessarily the whole space in which it is. $\endgroup$ – nicomezi Jan 29 '18 at 17:47
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$\text{ker}F$ is a space of matrices here, so you need to find a basis consisting of matrices for this subspace. First letting $b=1$ and $d=0$, then the opposite, we obtain the basis $$\left\{ \left[ {\begin{array}{cc} 2 & 1\\ 0 & 0\\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & 0\\ 2 & 1\\ \end{array} } \right]\right\}$$

The coordinate vector of the matrix in the question is the vector $(a,b)$ such that $\left[ {\begin{array}{cc} 2 & 1\\ 4 & 2\\ \end{array} } \right]=a\left[ {\begin{array}{cc} 2 & 1\\ 0 & 0\\ \end{array} } \right]+b\left[ {\begin{array}{cc} 0 & 0\\ 2 & 1\\ \end{array} } \right]$.

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  • $\begingroup$ can basis of a matrix be a collection of matrices? What I understood was that basis is the collection of vectors which are linearly independent and span the whole 'X' space. $\endgroup$ – Tumul Kumar Jan 29 '18 at 17:47
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    $\begingroup$ @TumulKumar Yes, a basis is exactly that. A collection is the same as a set by the way. $\endgroup$ – The Phenotype Jan 29 '18 at 17:49
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    $\begingroup$ @TumulKumar $Ker(T)$ is a subspace, in your case $Ker(T)\subseteq \mathbb{R}^{2\times 2}$ so you got a general representation of an element in $Ker(T)$ now you need to find a basis. "Pulling out" the scalars (This can be done due to $Ket(T)$ and matrices vector space gives you a spanning set which is minimal and therefore a basis $\endgroup$ – gbox Jan 29 '18 at 17:59
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You should have $Av=0$ and if $A$ is equal the following matrix:

$$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} $$

Then $a=2b$ and $ c=2d$. Now $A$ is the following one:

$$\begin{bmatrix} 2b & b \\ 2d & d \\ \end{bmatrix} $$

and its basis is:

$$\left\{\begin{bmatrix} 2 & 1 \\ 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 2 & 1 \\ \end{bmatrix}\right\}$$

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