Linear transformation between bases

[gothlev]

Hi !

I am a little bit confused with notation in the following:

Let A=

\begin{bmatrix}<br /> 2 &amp; 3 &amp; 4 \\<br /> 8 &amp; 5 &amp; 1 \\<br /> \end{bmatrix}

and consider A as a linear transformation mapping \mathbb{R}^3 to \mathbb{R}^2. Find the matix representation of A with respect to the bases

\begin{bmatrix}<br /> 1\\<br /> 1\\<br /> 0\\<br /> \end{bmatrix} , \begin{bmatrix}<br /> 0\\<br /> 1\\<br /> 1\\<br /> \end{bmatrix} , \begin{bmatrix}<br /> 1\\<br /> 0\\<br /> 1\\<br /> \end{bmatrix} of \mathbb{R}^3 and

\begin{bmatrix}<br /> 3\\<br /> 1\\<br /> \end{bmatrix} , \begin{bmatrix}<br /> 2\\<br /> 1\\<br /> \end{bmatrix} of \mathbb{R}^2

It seems to be a lot of A´s in here with different meanings, and I suppose it is what confuses me :(. Anyway I solved it as follows:

\begin{bmatrix}<br /> 3 &amp; 2\\<br /> 1 &amp; 1\\<br /> \end{bmatrix}^{-1} * \begin{bmatrix}<br /> 2 &amp; 3 &amp; 4\\<br /> 8 &amp; 5 &amp; 1\\<br /> \end{bmatrix} * \begin{bmatrix}<br /> 1 &amp; 0 &amp; 1\\<br /> 1 &amp; 1 &amp; 0\\<br /> 0 &amp; 1 &amp; 1\\<br /> \end{bmatrix} = \begin{bmatrix}<br /> -21 &amp; -5 &amp; -12\\<br /> 34 &amp; 11 &amp; 21\\<br /> \end{bmatrix}
I am still not sure that I have not confused myself with all the different A´s :( Am I on the right track or completely lost ?

 

[loveequation]

I think you are right but no guarantees. Just as a vector has different representations (components) in different bases, a matrix has different representations in different bases.

 

[HallsofIvy]

In general, to find the matrix representation of A, from U to V, with \{u_1, u_2, ..., u_n\} a basis for U and \{v_1, v_2, ..., v_m\} a basis for V:

Apply A to each of \{u_1, u_2, ..., u_n\} in turn and write the result as a linear combination of the \{v_1, v_2, ..., v_m\}. The coefficients form the columns of the matrix.

for example, here
u_1= \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}
so
Au_1= \begin{bmatrix}2 &amp; 3 &amp; 4 \\8 &amp; 5 &amp; 1 \\\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}5 \\ 13\end{bmatrix}

Now, [5, 13]= a[3, 1]+ b[2, 1] gives 3a+ 2b= 5 and a+ b= 13. Multiplying the second equation by 2, 2a+ 2b= 26 and, subtracting that from the first equation a= -21. -21+ b= 13 gives b= 34. The first column of the matrix you want is
\begin{bmatrix}-21 \\ 34\end{bmatrix}

 

[gothlev]

Thx for the replies. Thank you for a very clear and good explanation (HallsofIvy), the book I am reading is very compact and does not give very good explanations. There was a typo in the end of your reply:

\begin{bmatrix}-21 \\ 13\end{bmatrix}

should be

\begin{bmatrix}-21 \\ 34\end{bmatrix}

 

[HallsofIvy]

Thx for the replies. Thank you for a very clear and good explanation (HallsofIvy), the book I am reading is very compact and does not give very good explanations. There was a typo in the end of your reply:

\begin{bmatrix}-21 \\ 13\end{bmatrix}

should be

\begin{bmatrix}-21 \\ 34\end{bmatrix}

Right! Thanks.

No, I'll go back and edit my post so I can claim I never made that silly mistake!