Injective Linear transformation

Question

Given a linear transformation $T$ from $\mathbb{R}^4$ to $\mathbb{R}^4$

$$T(x,y,z,t) = (x-y+t,x-z,y-2z,2x-2z)$$

Is there an $n$ such that $T^n$ is injective?

I know that in order for a transformation to be injective it's kernel has to be equal to $\{0\}$.
Using the standard basis I could represent $T$ in a matrix fairly easy but probably it's not the right way to go about it since we are looking for a composition of $T$ with itself $n$ times

Answer 1

Note that a matrix $M$ is invertible if and only if its determinant is nonzero. Also note that if $M$ and $N$ are matrices, then $\det(MN) = \det(M)\cdot \det(N)$.

From this, it follows that $\det(T^n) = \det(T)^n$. Hence $\det(T^n)=0$ if and only if $\det(T)=0$, so $T^n$ is invertible if and only if $T$ is. Hence it suffices to find the determinant of $T$ as a matrix and check whether it's zero or not.

As a matrix, $$T = \begin{bmatrix}1 & -1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & -2 & 0 \\ 2 & 0 & -2 & 0\\\end{bmatrix}$$

We can prove that $T$ has a zero determinant and that therefore $T^n$ is not injective for any $n=1,2,3,\ldots$. With the benefit of hindsight, (see below!) it's enough to point out that

$$C_1 + 2C_2 + C_3 + C_4 = \vec{0}$$

where the $C_i$ denote the columns of $T$. Because this is a nonzero linear combination of columns that results in a zero column, the columns must be linearly-dependent. Hence the determinant of $T$ is zero. Hence the determinant of $T^n$ is zero for all $n\geq 1$. Hence $T^n$ isn't injective for any $n\geq 1$.

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Without such hindsight, we begin by trying to compute the determinant of $$T = \begin{bmatrix}1 & -1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & -2 & 0 \\ 2 & 0 & -2 & 0\\\end{bmatrix}$$

We perform elementary operations on this matrix to reduce it to a diagonal matrix whose determinant is easy to calculate.

• We exchange the first and last columns (flipping the sign of the determinant). $$\begin{bmatrix}1 & -1 & 0 & 1 \\ 0 & 0 & -1 & 1 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & -2 & 2\\\end{bmatrix}$$

• To the fourth column, we add the third column and subtract the first (no change in determinant). $$\begin{bmatrix}1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & -2 & 0\\\end{bmatrix}$$

• We normalize the last column by dividing by -2 (causes the determinant to be divided by -2). $$\begin{bmatrix}1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & -2 & 0\\\end{bmatrix}$$

• At this point, I note that to the first column, we can add the second and subtract the last. The result is a column of all zeros, which establishes the determinant as 0.

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

Answer 2

If $T$ is injective, just take $n=1$. Otherwise, $\det T=0$ and therefore $$(\forall n\in\mathbb{N}):\det(T^n)=(\det T)^n=0,$$which implies that $T^n$ is not injective, for each $n$.

So, all you have to do is to see whether $T$ is injective or not.