Proof of Linear Transformation Kernel

[user123321]

I have attempted part i) by creating a matrix X with random variable a, b, c and d, then performing the linear transformation on it. Then I made (F(X)|0), and attempted to solve linearly by back substitution. However, it just made a bigger mess than I started with. Am I on the right track or do I need to something different entirely? Thanks

 

[Deleted member 4993]

View attachment 21041

I have attempted part i) by creating a matrix X with random variable a, b, c and d, then performing the linear transformation on it. Then I made (F(X)|0), and attempted to solve linearly by back substitution. However, it just made a bigger mess than I started with. Am I on the right track or do I need to something different entirely? Thanks

Please share your "bigger mess" with us - so that we know where is the mistake and how we can help you.

 

[HallsofIvy]

Well, I wouldn't use "a, b, c, and d" because there is already an "a" in the equation! Let \(\displaystyle X= \begin{pmatrix}p & q \\ r & s \end{pmatrix}\) so that \(\displaystyle X^T= \begin{pmatrix}p & r \\ q & s \end{pmatrix}\). Then \(\displaystyle F(X)= X+ aX^T= \begin{pmatrix}p & q \\ r & s \end{pmatrix}+ \begin{pmatrix}ap & ar \\ aq & as \end{pmatrix}= \begin{pmatrix} p+ ap & q+ ar \\ r+ aq & s+ as\end{pmatrix}\).

Setting that equal to 0 we have p+ ap= p(1+ a)= 0, q+ ar= 0, r+ aq= 0, and s+ as= s(1+ a)= 0. From p(1+ a)= 0 and s(1+ a)= 0, either a= -1 or p= s= 0. If p= qs= 0 since q+ ar= 0, r= 0 and since r+ aq= 0, r= 0. Therefore, if a is not -1, \(\displaystyle X= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}\).