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Let $T:M_2(R) \to P_3$ be a linear transformation satisfying

$T\begin{bmatrix}1&0\\0&0\end{bmatrix} = x^3-2x$

$T\begin{bmatrix}0&-2\\1&0\end{bmatrix} = x^2-x+5$

$T\begin{bmatrix}0&-3\\0&1\end{bmatrix} = 5x^2+4x$

Find $T\begin{bmatrix}1&0\\3&-2\end{bmatrix}$

I can see that the first 3 matrices forms a basis of $M_2(R)$ but I don't know what to do next.

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  • 2
    $\begingroup$ Hint: write the last matrix as a linear combination of the first 3, then use the linearity of T. $\endgroup$ – Mahmoud Apr 3 '16 at 19:13
  • $\begingroup$ @Mahmoud I found the coefficients of the matrices which is a=1, b=3, c=-2, respectively. So would the answer be $1(x^3-2x)+3(x^2-x+5)-2(5x^2+4x)=x^3-7x^2-13x+15?$ $\endgroup$ – sucksatmath Apr 3 '16 at 19:31
  • $\begingroup$ exactly. because $T(ax+by+cz)=aT(x)+bT(y)+cT(z)$ for $a,b,c$ scalars and $x,y,z$ vectors (in this case = matrices), $\endgroup$ – Mahmoud Apr 3 '16 at 19:39

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