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Let $\space t:\mathbb{R^3} \to \mathbb{R^3}$ such that $\space t(v)=0 \space \forall \space v \in \mathbb{R^3}$. The matrix form would be a $3$ by $3$, zero matrix.

If $\space t=f \circ f$, where $f:\mathbb{R^3} \to \mathbb{R^3}$ and $f\neq t$, what is the matrix form of $f$?

I know that $\space t=f \circ f= f\cdot f$ and if one let $A$ to be the matrix of $f$ and $\space T$ the matrix of $t$, then $T=A^2$. And so $A^2$ would be a zero matrix.

Is there a systematic way to solve this problem? Thanks for the hints.

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2 Answers 2

up vote 4 down vote accepted

Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear transformation satisfying $T^2 = 0$.

First of all, let us check that the kernel of $T$ has dimension at least 2. Obviously the kernel cannot be trivial because that would mean $T$ is an isomorphism. Let us suppose that the kernel has dimension exactly 1, generated by a vector $v \in \mathbb{R}^3$. We can complete this to a basis $\{ v, v_1, v_2 \}$. Since $T^2 = 0$, $T(v_1) = \lambda_1 v$ and $T(v_2) = \lambda_2 v$ for some $\lambda_1, \lambda_2 \in \mathbb{R} \setminus \{0\}$. The problem is that $T(v_1-\frac{\lambda_1}{\lambda_2}v_2) = 0$ as well, even though $v_1-\frac{\lambda_1}{\lambda_2}v_2$ is not in the kernel! So the kernel has dimension at least 2.

If the dimension of the kernel is 3, then the matrix for $T$ in any basis is the zero matrix.

If the dimension of the kernel is 2, then matrices for $T$ are all similar to a matrix which is zero everywhere except for a 1 at the upper-right corner.

Why is this true? Since the kernel has dimension 2, we can find one vector $w$ which is not in the kernel. However, $T(w)$ is in the kernel. Now pick another vector $u$ in the kernel which is linearly independent from $T(w)$. With respect to the basis $\{ T(w), u, w \}$, the matrix for $T$ is the special matrix we singled out.

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If you multiply two $3 \times 3$ matrices together, the entries are the dot product of the first's rows with the second's columns. For the dot product to be zero, the rows/columns must be perpendicular. So, find a matrix where every column is perpendicular to every row (use zero rows and columns to make things easier).

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I see, I have to use the orthogonality spaces concept. But, does a space is orthogonal to itselfe? –  João May 27 '13 at 2:30
    
There is such a matrix for $n\ge 2$. For example, $\left[\begin{matrix}0 & \cdots & 1 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 0\end{matrix}\right]$, but I'm not sure how one would describe the entire set of matrices with that property. –  Henry Swanson May 27 '13 at 2:39
    
It seems that a matrix has the zero square iff all its eigenvalues are zero and its Jordan canonical form contains only the cells of order not greater than 2. –  Alex Ravsky Aug 21 '13 at 11:20

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