Question

Linear transformation A: R3 --> R3 is given as
A(a1)=(3-2k)*a1+(2-k)*a2+a3
A(a2)=(2-k)*a1+(2-k)*a2+a3
A(a3)=a1+a2+(2-k)*a3
where k is real, a1, a2, a3 are vectors of some basis of R3.
a) rank(A)=?
b) for k for which rank(A) is minimal, determine one basis of KerA and ImA.

I'm not sure about matrix of linear transformation. Is that matrix
(3-2k) (2-k) 1
B= (2-k) (2-k) 1
1 1 (2-k) ?
If yes, is the rank of lin. transf. A the same as the rank of that matrix B?
Is the way to find KerA this: B*x=0, where
x=[x1 x2 x3]' and 0=[0 0 0]'?

Answer 1

* What about ImA? Thanks

Answer 2

Whoa, what? Lol. It's so hard to read this question.

Answer 3

why is it hard?

Answer 4

abb0t is right. It's too hard for me to read this as well due to the limitations in this website. My advice will be to draw the matrix in mspaint and upload a picture of it

Answer 5

Try this ( yes, the Equation facility here takes a little learning, but saves time in the long run ) :
\[\left[\begin{matrix}(3-2k) & (2-k) &1\\ (2-k) & (2-k)&1\\1 & 1 & (2-k)\end{matrix}\right]\left[\begin{matrix}a_{1} \\a_{2}\\a_{3}\end{matrix}\right]\]so now you need to row reduce, keeping k as a parameter to vary later. I'd firstly permute to get a better first pivot :
\[\left[\begin{matrix}1 & 1&(2-k)\\ (3-2k) & (2-k)&1\\(2-k) & (2-k) & 1\end{matrix}\right]\left[\begin{matrix}a_{3} \\a_{1}\\a_{2}\end{matrix}\right]\]which I can do provided that I permute the vector components as well, so we can keep track of which component goes with which array row. To bring the A(2,1) position to zero then do row2 minus (3-2k) times row1, keeping in mind that :
\[(2-k)-(3-2k)= 2 -3 -k+2k=-1+k\]and \[1-(2-k)(3-2k)=1- (6-3k-4k+2k^{2}))= -5+7k-2k^{2}\]\[\left[\begin{matrix}1 & 1&(2-k)\\ 0 & (-1+k)&(-5+7k-2k^{2})\\(2-k) & (2-k) & 1\end{matrix}\right]\left[\begin{matrix}a_{3} \\a_{1}\\a_{2}\end{matrix}\right]\]yes, the algebra is painful, but continue with row3 minus (2-k) times row1 :
\[1 - (2-k)^{2} = 1 - (4 - 4k + k^{2})=\]\[-3 + 4k-k^{2}=-(k^{2}-4k+3)=\]\[-(k-1)(k-3)=(1-k)(k-3)\]\[\left[\begin{matrix}1 & 1&(2-k)\\ 0 & (-1+k)&(-5+7k-2k^{2})\\0 & 0 & (1-k)(k-3)\end{matrix}\right]\left[\begin{matrix}a_{3} \\a_{1}\\a_{2}\end{matrix}\right] \]When they ask for rank, they are asking how many columns ( or rows for that matter ) are independent? That depends on k, as the pivot values depend upon k. Inspection immediately gives you \[rank(A) = 3 \]\[k \neq 1, k\neq 3\]with\[rank(A)=2\]\[k \neq 1, k=3\]and\[rank(A) = 1\]\[k=1\]as these choices makes the pivots in A(2,2) and/or A(3,3) zero. Now to continue with part (b) take k = 1 ( that giving the minimal rank ):
\[\left[\begin{matrix}1 & 1&1\\ 0 &0&0\\0 & 0 & 0\end{matrix}\right]\left[\begin{matrix}a_{3} \\a_{1}\\a_{2}\end{matrix}\right]\]not forgetting to permute back to the original order\[\left[\begin{matrix}0 & 0&0\\ 0 &0&0\\1 & 1 & 1\end{matrix}\right]\left[\begin{matrix}a_{1} \\a_{2}\\a_{3}\end{matrix}\right]\]I must admit to not being familiar with notation KerA and ImA, but I'd guess that the column space of A ( with k = 1, t is real ) is\[t \left[\begin{matrix}0\\0\\ 1\end{matrix}\right]\]that is all the linear combinations of the only truly independent column. The null space is those vectors\[a=\left[\begin{matrix}a_{1} \\a_{2}\\a_{3}\end{matrix}\right]\]giving\[Aa = 0\]inspection gives\[\left[\begin{matrix}1 \\-1\\0\end{matrix}\right]\]and\[\left[\begin{matrix}1 \\0\\-1\end{matrix}\right]\]as a basis for that space ( but that's not the only choice you could make ) ie. all linear combinations thereof\[a = s\left[\begin{matrix}1 \\-1\\0\end{matrix}\right] + t\left[\begin{matrix}1 \\0\\-1\end{matrix}\right]\]with the null vector included by setting s and t to zero. Clever punters will note a relationship b/w the column space of A, the null space of A and the dimensions of A.

Answer 6

Sorry column space of A is \[t \left[\begin{matrix}1 \\1 \\ 1\end{matrix}\right]\]can anyone tell me what is meant by the notations KerA and ImA ??

Answer 7

\[KerA = \left\{ x | A(x)=0 \right\}\]
\[ImA = \left\{ x| \exists y, A(y)=x \right\}\]

Answer 8

Aha. KerA is thus the null space, ImA is the column space. 'Ker' is probably short for kernel, 'Im' is evidently image.