I have linear transformation $ \Bbb{R}^3\to \Bbb{R}^2 $, matrix of LT with respect to bases $B = \{(1;1;1),(0;1;0),(1;1;0)\}$ and $B'=\{(1;1),(0;1)\}$ is:

$$A = \begin{bmatrix}2 & 1 & 3\\3 & 1 & -3\end{bmatrix}$$

My task is to compute matrix $A'$ (LT with respect to standard bases), and using transition matrices. I suppose I have to compute transition matrix from std basis $S$ to basis $B$ in $ \Bbb{R}^3$, so $T_{SB}$ , then ${T'}_{B'S'}$ in $ \Bbb{R}^2$ and get the matrix $A'$ using formula $A'=T'\cdot A\cdot T$.

Is it correct? I am just not able to get correct solution this way.

  • $\begingroup$ Use change of the basis en.wikipedia.org/wiki/Change_of_basis. $\endgroup$
    – Arun
    Feb 1 '17 at 17:34
  • $\begingroup$ The set of vectors in B is not a basis. $\endgroup$
    – H. H. Rugh
    Feb 1 '17 at 17:43
  • $\begingroup$ @H.H.Rugh ofcourse it is! It is a set of linear independent vectors. It may not be orthogonal, but it is a basis! $\endgroup$
    – Laray
    Feb 1 '17 at 17:45
  • $\begingroup$ @Laray I've fixed it, it really wasn't a basis before. $\endgroup$
    – Matt
    Feb 1 '17 at 17:46
  • $\begingroup$ Yes, now it is. $\endgroup$
    – H. H. Rugh
    Feb 1 '17 at 17:46

Your procedure seems correct, but it depends on how you write down these matrices.

Let $(e_i)_{i=1,2,3}$ and $(e'_j)_{j=1,2}$ denote the canonical bases in the source and the image. Let $B=(u_\ell)_{\ell=1,2,3}$ and $B'=(u'_k)_{k=1,2}$ denote the stated bases vectors and finally $A=(a_{kl})$ the matrix calculated in the bases $B,B'$. (As a paranthetical remark: One of my predelictions concerning change of base calculations is to write an abstract vector $x$ in the form $x=\sum_k u_k x_k$ in the basis $B$, i.e. with the vectors to the right. This ensures keeping correct track of indices and matrix multiplications. Paranthesis closed).

The abstract linear transformation acts as follows: $$ f(u_\ell) = \sum_{k=1}^2 u'_k A_{k\ell} $$ Writing $u_\ell= \sum_{i=1}^3 e_i u_{i\ell}$ and $u'_k= \sum_{j=1}^2 e'_j u'_{jk}$ we get the identity: $$ \sum_i f(e_i) u_{i\ell} = \sum_j e'_j \sum_k u'_{jk} A_{k\ell} $$ Let $M$ be the matrix of $f$ between the canonical bases. Then $$ \sum_i f(e_i) = \sum_j e'_j M_{ji}$$ So comparing we get: $\sum_i M_{ji} u_{i\ell} = \sum_k u'_{jk} A_{k\ell}$ or with your data:

$$ M \left[\begin{matrix} 1 & 0 & 1\\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} \right]= \left[\begin{matrix} 1 & 0\\ 1 & 1 \end{matrix} \right] \left[\begin{matrix} 2 & 1 & 3\\ 3 & 1 & -3 \end{matrix} \right] $$ which you then have to solve for $M$. The answer is then simply $M'$ the usual transpose of $M$. Using scilab I got:$$ M= \left[\begin{matrix} 2 & 1 & -1\\ -2 & 2 & 5 \end{matrix} \right]$$ so the answer should be its transpose. You may now tell me if this is compatible with the answer to the exercise?

  • $\begingroup$ Yes, thank you for explanation. $\endgroup$
    – Matt
    Feb 3 '17 at 10:06

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