# How to find $\text {Ker} (T)$ when you do not have a matrix $A$ that represents the transformation.

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.

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

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.

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 $$ker (T) = \{0 \}.$$

• $\ker(T)$ does not contain all polynomials of degree$\ge3$ Mar 28, 2019 at 17:22
• Because constant coefficient, coefficient of $x$ and coefficient of $x^2$ are all zero by the condition that $p(0)=p'(0)=p''(0) =0.$ Mar 28, 2019 at 17:23
• $x^3+x$ is also a polynomial of degree $3$ but doesn't lie in the kernel. Perhaps you meant all polynomials with the lowest exponent of $x\ge3$? Mar 28, 2019 at 17:24
• 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. Mar 28, 2019 at 17:26
• Oh! Yeah. Sorry. I will edit. Mar 28, 2019 at 17:27

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 $$T$$.

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 $$c$$.