Linear transformation

Question

Let there be a linear transformation $T:R^3\rightarrow R^2$

Is there a linear transformation so that:

$Ker(T)=Span((1,2,1),(0,3,-1))$ and $Im(T)=Span((5,-7))$

Answer: $Dim(V)=Rank(T)+Null(T)=2+1=3$ so by dimension there is now to find $T$

$\begin{pmatrix} a&b&c\\ d&e&f\\ \end{pmatrix}* \begin{pmatrix} 1\\ 2\\ 1 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\end{pmatrix}$

$\begin{pmatrix} a&b&c\\ d&e&f\\ \end{pmatrix}*\begin{pmatrix} 0\\ 3\\ -1 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

So I got $\begin{pmatrix} -5b &b &3b\\ -5e&e&3e\\ \end{pmatrix}*\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}=\begin{pmatrix} 5\\ -7\\ \end{pmatrix}$

How do I continue from here?

Answer 1

The simple Theorem to remember is that given a basis $\{\alpha_1, \alpha_2, ..., \alpha_n\}$ of $V$ and any $n$ vectors $\{\beta_1, \beta_2,.., \beta_n, \}$ in $W$ there is exactly one Linear Transformation such that $ T(\alpha_i) = \beta_i$. So you have two simple tasks:

• Find a basis $\{v_1, v_2\} $for $\ker T$ and let $T(v_1) = T(v_2) = \underline {0}$
• You can extend $\{v_1, v_2\}$ by adding one more vector $v_3$ to form a basis for $\Bbb R^3$. Define $T(v_3) = (5, -7)$

This $T$ is uniquely determined and conforms to your conditions.

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Now I get the feeling you also require the matrix which represents this particular Linear Transformation. The matrix of $T$ relative to the bases $ \{v_1, v_2, v_3\} $ and $\{(0,1), (1, 0)\}$is: $$ \left({\begin{matrix} 0 & 0 & 5 \\ 0 & 0 & -7 \\ \end{matrix}}\right)$$

which you can transform to a matrix in the Standard Ordered Bases using the theory of similar matrices (see pages 91 - 94). And this will also allow you to find an explicit general equation for $T(x, y, z)$

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UPDATE:

This thread is too long so here is a proper solution. Note that adding another vector would make $\{(1,2,1), (0,3,-1), (0,0,1)\}$ would create a basis (which can be verified by row reduction). Now, as specified above, we define $T$ thus,

• $ T(1,2,1) = (0,0) $
• $ T(0,3,-1) = (0,0) $
• $ T(0,0,1) = (5,-7) $

Such a definition is valid according to the Theorem I have stated above and linked first. Now since the three vectors form a basis there are uniquescalars $a, b, c$ such that $$ (x, y, z) = a(1,2,1) + b(0,3,-1) + c(0,0,1) \;\; \forall (x, y, z) \in \Bbb R^3 $$

The scalars can be found by solving a simple system of equations.

Now for any $(x, y, z)$ we have,

$$T(x, y, z) = T(a(1,2,1) + b(0,3,-1) + c(0,0,1)) = a T(1,2,1) + b T(0,3,-1) + c T (0,0,1) $$

Substitute for the scalars and $Tv$'s and you would have an explicit formula for $T$ in terms of $ x, y, z $.