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Usually finding $\ker(T)$ with a matrix that represents the transformation is relatively straightforward. You solve for $Ax = 0$ by using gauss-jordan.

But when you do not have a matrix that represents the transformation (it's not provided), solving $Ax = 0$ is rather confusing and less straightforward. I want to know if there is a general pattern or guidelines in solving this types of problems.

Examples:

$1)$ $T: \Bbb M_{2 \times 2} \longrightarrow P_2[x]$ defined by $$T\left [\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right ] \mapsto (a+b) + (b+c)x + (c+d)x^2.$$

$2)$ $T: P_2[t] \longrightarrow \Bbb M_{2 \times 2}$ defined by $$p(t) \mapsto \begin{pmatrix} p(0) & p'(0) \\ p'(0) & p''(0) \end{pmatrix}.$$

I am not asking how to solve these to examples, I am asking what are the guidelines (Steps) to solve problems of the same genre.

If you could also point me two a ressource where I could find more problems of this type so I can practice myself; I would be really grateful.

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  • $\begingroup$ For the first problem $ker (T) = \left \{\begin{pmatrix} a & -a \\ a & -a \end{pmatrix} \in \Bbb M_{2 \times 2} (\Bbb R) : a \in \Bbb R \right \}.$ $\endgroup$
    – little o
    Mar 28, 2019 at 17:06

4 Answers 4

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The first comes down to finding all matrices $\begin{pmatrix}a&b\\c&d \end{pmatrix}$ such that its $T$-value is the $0$-polynomial $0+0\cdot x+0\cdot x^2$, i.e. $a+b=0$ and $b+c=0$ and $c+d=0$. This is just a standard linear system of 3 equations in $4$ variables. It consists of all matrices of the form (using $d$ as a free variable) $\begin{pmatrix}-t & t \\ -t & t\end{pmatrix}, t \in \mathbb{R}$.

The last looks for all polynomials $p(t)$ of at most degree $2$ such that $p(0)=0=p'(0)=p''(0)$ and setting $p(t)=a+bt + ct^2$, so $p'(t)=b+2ct$, $p''(t)=2c$ these conditions yield $2c=0$ so $c=0$, then using the $p'(0)=0$ we get $b=0$ and finally $p(0)=0$ gives $a=0$, so only the $0$-polynomial lies in the kernel.

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For the first problem $$ker (T) = \left \{\begin{pmatrix} a & -a \\ a & -a \end{pmatrix} \in \Bbb M_{2 \times 2} (\Bbb R) : a \in \Bbb R \right \}$$ and for the second problem $ker(T)$ contains all the polynomials of which has no term upto $x^2.$ But the degree of each polynomial in $\Bbb P_2[t]$ is at most $2.$ So for the second problem $ker (T) = \{0 \}.$

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    $\begingroup$ $\ker(T)$ does not contain all polynomials of degree$\ge3$ $\endgroup$ Mar 28, 2019 at 17:22
  • $\begingroup$ Because constant coefficient, coefficient of $x$ and coefficient of $x^2$ are all zero by the condition that $p(0)=p'(0)=p''(0) =0.$ $\endgroup$
    – little o
    Mar 28, 2019 at 17:23
  • $\begingroup$ $x^3+x$ is also a polynomial of degree $3$ but doesn't lie in the kernel. Perhaps you meant all polynomials with the lowest exponent of $x\ge3$? $\endgroup$ Mar 28, 2019 at 17:24
  • $\begingroup$ But the problem is that there is no polynomial in $\Bbb P_2[t]$ of degree greater or equal to $3.$ So the only polynomial satisfying the result is the zero polynomial. $\endgroup$
    – little o
    Mar 28, 2019 at 17:26
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    $\begingroup$ Oh! Yeah. Sorry. I will edit. $\endgroup$
    – little o
    Mar 28, 2019 at 17:27
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You always have matrices representing the linear map. Let's do the first one.

Consider the basis $$ E_1=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad E_2=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad E_3=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \quad E_4=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $$ of the domain and the basis $\{1,x,x^2\}$ of the codomain.

Then the matrix of $T$ with respect to these bases is $$ \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix} $$ and the RREF is $$ \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \end{bmatrix} $$ The null space of this matrix has a basis consisting of the single vector $$ \begin{bmatrix} -1 \\ 1 \\ -1 \\ 1 \end{bmatrix} $$ so that the matrix this is the coordinate vector of, namely $$ -E_1+E_2-E_3+E_4=\begin{bmatrix} -1 & 1 \\ -1 & 1 \end{bmatrix} $$ forms a basis of the kernel of $T$.

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When in doubt, fall back on basic definitions: the kernel of a linear transformation is the set of vectors in its domain that get mapped to $0$.

So, for the first problem, you’re looking for $a,b,c,d\in\mathbb R$ such that $(a+b)+(b+c)x+(c+d)x^2=0$ for all $x$. This means that all of the coefficients of this polynomial must vanish, which gives you a system of linear equations to solve for $a$, $b$, $c$ and $d$.

In the same vein, for the second problem, you’re looking for polynomials $p(t)=a+bt+ct^2$ such that $p(0)=p'(0)=p''(0)=0$. Each of these conditions generates a rather simple linear equation in $a$, $b$ and $c$.

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