Let $A$ be a complex matrix. The conjugate transpose of $A,$ denoted $A^*,$ is given by $\left(\overline{A} \right)^T,$ where $\overline{A}$ is the matrix whose entries are complex conjugates of the corresponding entries of $A.$

Note that if $A$ is real, $A^* = A^T.$

Let $A=\left(\begin{array}{cc}3+7i&0\\2i&4-i\end{array}\right).$ Write down the conjugate transpose of $A$.

Here we have that \[ \overline{A} = \left( \begin{array}{cc} 3-7i&0\\ -2i&4+i \end{array} \right). \]

and

Let $A=\left(\begin{array}{cc}3+7i&0\\2i&4-i\end{array}\right).$ Write down the conjugate transpose of $A$.

\[ \overline{A} = \left( \begin{array}{cc} 3-7i&0\\ -2i&4+i \end{array} \right), \qquad \ds A^* =\left(\overline{A}\right)^T = \left( \begin{array}{cc} 3-7i&-2i\\ 0&4+i \end{array} \right). \]

Also we have \[ {A}^T = \left( \begin{array}{cc} 3+7i&2i\\ 0&4-i \end{array} \right). \]

and \[ \overline{{A}^T} = \left( \begin{array}{cc} 3-7i&-2i\\ 0&4+i \end{array} \right). \]

Let $A=\left(\begin{array}{cc}3+7i&0\\2i&4-i\end{array}\right).$ Write down the conjugate transpose of $A$.

\[ \overline{A} = \left( \begin{array}{cc} 3-7i&0\\ -2i&4+i \end{array} \right), \qquad \ds A^* =\left(\overline{A}\right)^T = \left( \begin{array}{cc} 3-7i&-2i\\ 0&4+i \end{array} \right), \]

\[ {A}^T = \left( \begin{array}{cc} 3+7i&2i\\ 0&4-i \end{array} \right), \qquad \ds \overline{{A}^T} = \left( \begin{array}{cc} 3-7i&-2i\\ 0&4+i \end{array} \right). \]

In general \[ \left(\overline{A}\right)^T = \overline{{A}^T}. \]

Recall that in $\mathbb{R}^n$ the inner (or dot) product of two vectors $$ \u = \left( \begin{array}{c} u_1\\ u_2\\ \vdots\\ u_n\end{array}\right),\quad \v = \left( \begin{array}{c} v_1\\ v_2\\ \vdots\\ v_n\end{array}\right) $$ is given by $ \u\pd\v = u_1v_1 + u_2v_2 + \cdots + u_nv_n $ and the length (a real number!) of $\u$ by $$ ||\u|| = \sqrt{\u\pd\u} = \sqrt{u_1^2 + u_2^2+\cdots +u_n^2}. $$

Recall that in $\mathbb{R}^n$ the inner (or dot) product of two vectors $$ \u = \left( \begin{array}{c} u_1\\ u_2\\ \vdots\\ u_n\end{array}\right),\quad \v = \left( \begin{array}{c} v_1\\ v_2\\ \vdots\\ v_n\end{array}\right) $$ is given by $ \u\pd\v = u_1v_1 + u_2v_2 + \cdots + u_nv_n $ and the length (a real number!) of $\u$ by $$ ||\u|| = \sqrt{\u\pd\u} = \sqrt{u_1^2 + u_2^2+\cdots +u_n^2}. $$

These definitions are unsuitable for vectors in $\mathbb{C}^n$.

To demonstrate, consider the vector $\u = (i,1)$ in $\mathbb{C}^2$. Using the above expression for length, we would obtain $||\u|| = \sqrt{i^2+1} = 0,$ so $\u$ would be a non-zero vector with length $0$.

Instead, we introduce the complex inner product $$ \u\pd\v = u_1\overline{v}_1 + u_2\overline{v}_2 + \cdots + u_n\overline{v}_n, $$ where as usual $\overline{v}_i$ denotes the complex conjugate of $v_i$.

In matrix notation, we can write this as $\u\pd\v = \v^*\u$. Note the length of a complex vector is always a real number.

1) $\u\pd\v = u_1\overline{v}_1 + u_2\overline{v}_2 + \cdots + u_n\overline{v}_n= \overline{\v\pd \u}$

2) The length of a complex vector is always a real number.

Recall that for every $z_1,z_2$ complex objects, $\overline{z_1z_2} = \overline{z_1}~\overline{z_2},$ and $\overline{z_1+z_2} = \overline{z_1}+ \overline{z_2}.$

So now we understand what is meant by the following statement:

Columns of a unitary matrix form an orthonormal set with respect to the complex inner product.

A complex matrix $A$ is called Hermitian (or self-adjoint) if $A=A^*.$

As with symmetric matrices, we can recognise a Hermitian matrix by inspection. See if you can see the pattern in the following $2\times 2,$ $3\times 3$ and $4\times 4$ Hermitian matrices.

A complex matrix $A$ is called Hermitian (or self-adjoint) if $A=A^*.$

As with symmetric matrices, we can recognise a Hermitian matrix by inspection. See if you can see the pattern in the following $2\times 2,$ $3\times 3$ and $4\times 4$ Hermitian matrices.

A complex matrix $A$ is called Hermitian (or self-adjoint) if $A=A^*$.

As with symmetric matrices, we can recognise a Hermitian matrix by inspection. See if you can see the pattern in the following $2\times 2$, $3\times 3$ and $4\times 4$ Hermitian matrices.

One of the most significant results on Hermitian matrices is that their eigenvalues are real.

Let $\v\in \mathbb{C}^n$ be an eigenvector of the Hermitian matrix $A$, with corresponding eigenvalue $\lambda$. In other words, \begin{equation} A\v = \lambda \v. \qquad \quad (11) \end{equation} In what follows, we use the fact that $(AB)^* = B^*A^*$ which holds since the same is true for matrix transposition.

We multiply (11) from the left by $\v^*$ (treat $\v$ as an $n\times 1$ complex matrix) to obtain \begin{equation} \v^*A\v =\v^*(\lambda\v) = \lambda(\v^*\v). \qquad \qquad (12) \end{equation}

Let $\v\in \mathbb{C}^n$ be an eigenvector of the Hermitian matrix $A$, with corresponding eigenvalue $\lambda$. In other words, \begin{equation} A\v = \lambda \v. \qquad \quad (11) \end{equation} In what follows, we use the fact that $(AB)^* = B^*A^*$ which holds since the same is true for matrix transposition.

We multiply (11) from the left by $\v^*$ (treat $\v$ as an $n\times 1$ complex matrix) to obtain \begin{equation} \v^*A\v =\v^*(\lambda\v) = \lambda(\v^*\v). \qquad \qquad (12) \end{equation}

Also note that $$ (\v^*A\v)^* = \v^*A^*(\v^*)^* = \v^*A\v, $$ which yields, by applying (12), $$ {\rm l.h.s.}=(\lambda \v^*\v)^*=\bar{\lambda}\v^*\v={\rm r.h.s.}=\lambda\v^*\v. $$

Also note that $$ (\v^*A\v)^* = \v^*A^*(\v^*)^* = \v^*A\v, $$ which yields, by applying (12), $$ {\rm l.h.s.}=(\lambda \v^*\v)^*=\bar{\lambda}\v^*\v={\rm r.h.s.}=\lambda\v^*\v. $$

It follows that $$ (\bar{\lambda}-\lambda)\v^*\v=0,~~~{\rm so}~\bar{\lambda}=\lambda~({\rm as}~\v^*\v\neq 0), $$ i.e. $\lambda$ is real.

Also note that $$ (\v^*A\v)^* = \v^*A^*(\v^*)^* = \v^*A\v, $$ which yields, by applying (12), $$ {\rm l.h.s.}=(\lambda \v^*\v)^*=\bar{\lambda}\v^*\v={\rm r.h.s.}=\lambda\v^*\v. $$

It follows that $$ (\bar{\lambda}-\lambda)\v^*\v=0,~~~{\rm so}~\bar{\lambda}=\lambda~({\rm as}~\v^*\v\neq 0), $$ i.e. $\lambda$ is real.

One consequence of this result is that a real symmetric matrix has real eigenvalues, since every real symmetric matrix is Hermitian. This result was stated on Chapter 18 but not proved.

A square complex matrix is called normal if it commutes with its own conjugate transpose, ie, if $AA^* = A^*A$.

Normal matrices are generally more difficult to identify by inspection. However, we have some classes of matrices which are normal:

• unitary,
• Hermitian,
• real skew-symmetric (satisfying $A^T=-A$),
• any diagonal matrix,
• others?

A square complex matrix is called normal if it commutes with its own conjugate transpose, ie, if $AA^* = A^*A$.

Normal matrices are generally more difficult to identify by inspection. However, we have some classes of matrices which are normal:

• unitary,
• Hermitian,
• real skew-symmetric (satisfying $A^T=-A$),
• any diagonal matrix,
• others?

We make a note that real normal $2\times 2$ matrices are either symmetric or of the form $\mat{cc}{a & b\\ -b & a}$ (which include the skew-symmetric examples).

A class of matrix which is not generally normal is the class of complex symmetric matrices.

Classify the matrix $A=\mat{cc}{1&1+i\\ 1+i&-i}$.

$A^* = \left( \begin{array}{cc} 1 & 1-i \\ 1-i & i \end{array} \right) $ $\neq A.$ Then $A$ is not Hermitian.

$A^* A = \left( \begin{array}{cc} 1 & 1-i \\ 1-i & i \end{array} \right) \left( \begin{array}{cc} 1 & 1+i \\ 1-i & -i \end{array} \right) $ $= \left( \begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array} \right) $ $= 3I .$
Then $A$ is not unitary.

Similarly $A~A^* = 3I $ $=A^*A.$ Then $A$ is normal.

The main result we have is completely analagous to the real case of orthogonal diagonalisation and symmetric matrices on Chapter 18. We will not prove this result.

If possible, diagonalise the matrix $\mat{cc}{6 & 2+2i\\ 2-2i& 4}.$ 📝

Step 1. Determine whether or not the given matrix is unitarily diagonalisable. That is, check that $A^* = A.$

Step 2. Find eigenvalues $\lambda_1, \lambda_2$ and corresponding eigenvectors $\v_1,\v_2.$

Step 3. Form your unitary matrix $P = \big(\hat{\v}_1 ~|~ \hat{\v}_2\big).$ Check that $P^* P = I = P P^*.$

Step 4. Compute the diagonal matrix $D = P^*A P.$