# Thread: [SOLVED] Prove that a matrix that is both unitary and upper triangular must be diagon

1. ## [SOLVED] Prove that a matrix that is both unitary and upper triangular must be diagon

Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.

In case you don't remember what a unitary matrix is, please refer to wikipedia.

Possibly relevant theorem from the textbook:

Theorem 6.18. Let $\mathsf{T}$ be a linear operator on a finite-dimensional inner product space $\mathsf{V}$. Then the following statements are equivalent:

(a) $\mathsf{T}\mathsf{T}^*=\mathsf{T}^*\mathsf{T}=\mat hsf{I}$.

(b) $\langle\mathsf{T}(x),\mathsf{T}(y)\rangle=\langle x,y\rangle$ for all $x,y\in\mathsf{V}$.

(c) If $\beta$ is an orthonormal basis for $\mathsf{V}$, then $\mathsf{T}(\beta)$ is an orthonormal basis for $\mathsf{V}$.

(d) There exists an orthonormal basis $\beta$ for $\mathsf{V}$ such that $\mathsf{T}(\beta)$ is an orthonormal basis for $\mathsf{V}$.

(e) $\lVert\mathsf{T}(x)\rVert=\lVert x\rVert$ for all $x\in\mathsf{V}$.

Every time I start into this, it turns into something horribly messy. I feel like there's a trick I'm missing. Any thoughts? Or perhaps a full solution?

Thanks!

2. Originally Posted by hatsoff
Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.

In case you don't remember what a unitary matrix is, please refer to wikipedia.

Possibly relevant theorem from the textbook:

Theorem 6.18. Let $\mathsf{T}$ be a linear operator on a finite-dimensional inner product space $\mathsf{V}$. Then the following statements are equivalent:

(a) $\mathsf{T}\mathsf{T}^*=\mathsf{T}^*\mathsf{T}=\mat hsf{I}$.

(b) $\langle\mathsf{T}(x),\mathsf{T}(y)\rangle=\langle x,y\rangle$ for all $x,y\in\mathsf{V}$.

(c) If $\beta$ is an orthonormal basis for $\mathsf{V}$, then $\mathsf{T}(\beta)$ is an orthonormal basis for $\mathsf{V}$.

(d) There exists an orthonormal basis $\beta$ for $\mathsf{V}$ such that $\mathsf{T}(\beta)$ is an orthonormal basis for $\mathsf{V}$.

(e) $\lVert\mathsf{T}(x)\rVert=\lVert x\rVert$ for all $x\in\mathsf{V}$.

Every time I start into this, it turns into something horribly messy. I feel like there's a trick I'm missing. Any thoughts? Or perhaps a full solution?

Thanks!
Let $X$ be upper triangular and unitary, partition $X$ so that:

$X=\left[
\begin{array}{cc}A&B\\0&C\end{array}
\right]$

Where $A$ and $C$ are upper triangular square unitary matrices (not necessarily the same size).

Now consider:

$X^HX$

CB

3. So let's make sure I understand the proof correctly...

Choose $X$ an upper-triangular, unitary matrix. Partition $X$ so that

$X=\left[\begin{array}{cc}A&B\\0&C\end{array}\right]$,

with $A,C$ upper-triangular unitary matrices. Then we have

$XX^*=\left[\begin{array}{cc}A&B\\0&C\end{array}\right]\left[\begin{array}{cc}A^*&0\\B^*&C^*\end{array}\right]$

$=\left(\left[\begin{array}{cc}A&0\\0&C\end{array}\right]+\left[\begin{array}{cc}0&B\\0&0\end{array}\right]\right)\left(\left[\begin{array}{cc}A^*&0\\0&C^*\end{array}\right]+\left[\begin{array}{cc}0&0\\B^*&0\end{array}\right]\right)$

$=\left[\begin{array}{cc}A&0\\0&C\end{array}\right]\left[\begin{array}{cc}A^*&0\\0&C^*\end{array}\right]+\left[\begin{array}{cc}0&B\\0&0\end{array}\right]\left[\begin{array}{cc}A^*&0\\0&C^*\end{array}\right]$ $+\left[\begin{array}{cc}A&0\\0&C\end{array}\right]\left[\begin{array}{cc}0&0\\B^*&0\end{array}\right]+\left[\begin{array}{cc}0&B\\0&0\end{array}\right]\left[\begin{array}{cc}0&0\\B^*&0\end{array}\right]$

$=\left[\begin{array}{cc}AA^*&0\\0&CC^*\end{array}\right]+\left[\begin{array}{cc}0&BC^*\\0&0\end{array}\right]+\left[\begin{array}{cc}0&0\\CB^*&0\end{array}\right]+\left[\begin{array}{cc}BB^*&0\\0&0\end{array}\right]$

$=\left[\begin{array}{cc}AA^*+BB^*&BC*\\CB^*&CC^*\end{arra y}\right]=I_n$.

So $BC^*=CB^*=0$ $\Rightarrow$ $BC^*C=B=0$.

First of all, is that correct so far?

If yes, then I have one remaining question: Why can we assume $A$ and $C$ are unitary? Obviously they are upper-triangular, but how do I prove that if I arbitrarily partition $X$, we have $A,C$ unitary?

Thanks!

4. let $A=(a_{ij})$ be an $n \times n$ unitary and upper-triangular matrix. since $A$ is upper triangular, we have $\det A = a_{11}a_{22} \cdots a_{nn} \neq 0.$ thus $a_{ii} \neq 0, \ \forall i.$ now let $e_j \in \mathbb{C}^n$ be the column with $1$ in the $j$-th

entry and 0 everywhere else. note that $Ae_j$ is the $j$-th column of $A.$ let $1 < j \leq n.$ then $0===a_{11}a_{1j},$ and thus $a_{1j}=0,$ for all $1 < j \leq n.$ so all the entries in the first

row of $A$ except $a_{11}$ are 0. next, let $2 < j \leq n.$ then $0===a_{22}a_{2j},$ and thus $a_{2j}=0,$ for all $2 < j \leq n.$ so all the entries in the second row of $A$ except $a_{22}$ are 0.

continuing this way we see that for any $1 \leq i \leq n,$ the only non-zero entry in the $i$-th row of $A$ is $a_{ii},$ and hence $A$ is diagonal.

5. Originally Posted by hatsoff
If yes, then I have one remaining question: Why can we assume $A$ and $C$ are unitary? Obviously they are upper-triangular, but how do I prove that if I arbitrarily partition $X$, we have $A,C$ unitary?

Thanks!
$X^H=X=\left[\begin{array}{cc}A^H&0\\B^H&C^H\end{array}\right]$

and:

$X^HX=\left[\begin{array}{cc}A^H&0\\B^H&C^H\end{array}\right]\left[\begin{array}{cc}A&B\\0&C\end{array}\right]$ $=\left[\begin{array}{cc}A^HA&A^HB\\B^HA&B^HB+C^HC\end{arr ay}\right]=I_{sz(X)}$

so:

$A^HA=I_{sz(A)}$

and to show $CC^H=I_{sz(C)}$ we need only consider $XX^H$

It might be easier to see what is going on if you let $C$ be a $1 \times 1$ matrix (and this is all that is needed for the obvious induction required to complete the proof by this method)

CB

6. Originally Posted by hatsoff
Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.
Alternative:

Let the matrix be $A$.

If you know that 'the inverse of an upper triangular matrix is upper triangular', the proof is simpler.

Inverse of unitary $A$ is $A^H$. But $A^H$ will be lower triangular. Since 'the inverse of an upper triangular matrix is upper triangular', $A^H$ must be upper triangular too. Thus $A^H$ is both upper triangular and lower triangular, thereby diagonal. So is $A$

7. Thanks a lot! This is all very helpful.

You guys really came through for me. I hope I can return the favor soon.

8. oh!! this post was really helpful. Thank you guys

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