Find (if exists) linear transformation.

[em_agda]

[MATH] \mathbb{V}=\mathrm{span}\left(\left[ \begin{array}{cc} 1 & 0 \\ 1 & 1\end{array}\right], \left[ \begin{array}{cc} 1 & -1 \\ -1 & 1\end{array}\right]\right)[/MATH], [MATH]\eta: M_{2\times 2}(\mathbb{R})\to M_{2\times 2}(\mathbb{R})[/MATH] and [MATH] \mathrm{ker}\eta=\mathrm{im}\eta=\mathbb{V}.[/MATH] Find (if exist) [MATH] eta. [/MATH]

 

[Steven G]

What is eta?

 

[em_agda]

linear transformation

 

[em_agda]

eta is [MATH]\eta[/MATH]

 

[Steven G]

OK, thanks. So where are you stuck? What have you tried?

What is [MATH]\eta[/MATH] of each of the span vectors?

 

[em_agda]

So I thought that [MATH]\eta(x)=[0,0,0,0] [/MATH] (transposed) and I wrote that if v in V [MATH]v=\alpha [1,0,1,1]+ \beta[1,1,-1,1][/MATH], and dimension of V is 2, and matrix 2x2 dimension is 4, so ker and image dimension should be also 2 and I really don't know where I m going with this solution.

 

[Steven G]

OK, for some reason I did not see that you were asked to find η and I hoping that you were going to realize that you needed the definition of η.

So the kerη=imη=V. What does that mean. kerη=imη implies that every vector in V gets mapped to 0_V. Now this all fine and possible. But we are told that V = kerη=imη. So V = the 2x2 0-matrix for η to exist. But this is not possible as V contains the two non-zero matrices given that make up the span of V. Can you finish up from here?