give answer in 2 step with explanation at the end of each step and final answer at the end:9. Let M,,...(R) be the set of matrices of size m x n with real
coefficients. The operator norm of a matrix

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:9. Let M,,...(R) be the set of matrices of size m x n with real
coefficients. The operator norm of a matrix 4 € M,,....(R) is defined as
[All := sup{| Az] : x € R", |=] = 1}

Prove that:

(a) If A € My, (R), then [A] < oc,

(b) If A € My, (R), then 4] = 0 and 14] = Of and only if A = 0.

(©) If A, BE M,,,,(R), then |A + BJ < |] + | BI,

(d) If c € R, then leAl = |c|lAl.

(e) If AeM,.,(R)andze R", then lAz] < |All],

(A) If AEM, (R), BE M,,,(R), then |AB] < |All B].

(9) If A € M,,...(R), then | 4] is the square root of the largest eigenvalue
of the symmetric matrix A” A.

VIPSolutions · Accepted Answer

Let's work through part (a):

### (a) If $A \in M_{m \times n}(\mathbb{R})$, then $\|A\| < \infty$.

---
#### **Step 1:**  
Recall the definition:  
$$
\|A\| = \sup \{ \|Ax\| : x \in \mathbb{R}^n, \|x\| = 1 \}
$$
The value $\|Ax\|$ is a continuous function of $x$ on the compact set $\{x \in \mathbb{R}^n : \|x\| = 1\}$ (the unit sphere).

**Explanation:**  
The image of a compact set under a continuous function is also compact, hence bounded above.

---
#### **Step 2:**  
Since every entry of $A$ is real and finite, for any $x$ with $\|x\| = 1$,  
$$
\|Ax\|^2 = \sum_{i=1}^m \left( \sum_{j=1}^n a_{ij} x_j \right)^2
$$
Each term is bounded since both $a_{ij}$ and $x_j$ are bounded.

**Explanation:**  
Therefore, the supremum over all such $x$ is finite.

---
**Final Answer:**  
$\|A\| < \infty$ for all $A \in M_{m \times n}(\mathbb{R})$.

VIPSolutions · Answer

Let's proceed step-by-step with each part:

---

### (b) If $A \in M_{m \times n}(\mathbb{R})$, then $\|A\| = 0$ and $\|A\| = \text{some scalar}$ if and only if $A = 0$.

**Step 1:**  
Recall the definition of the operator norm:  
$$
\|A\| = \sup_{\|x\|=1} \|A x\|
$$  
If $A = 0$, then for all $x$, $A x = 0$, so $\|A x\| = 0$.  
Hence, $\|A\| = 0$.

**Explanation:**  
The zero matrix maps every vector to the zero vector, so its norm is zero.

---

**Step 2:**  
Conversely, suppose $\|A\| = 0$. Then, for all $x$ with $\|x\|=1$, $\|A x\| = 0$. By homogeneity, for arbitrary $x \neq 0$,  
$$
A x = \|x\| A \left( \frac{x}{\|x\|} \right) = \|x\| \cdot 0 = 0
$$  
Thus, $A x = 0$ for all $x$, which implies $A$ is the zero matrix.

**Final answer:**  
$\boxed{\text{The norm is zero if and only if } A = 0}$.

---

### (c) Triangle inequality: $\|A + B\| \leq \|A\| + \|B\|$.

**Step 1:**  
By definition,  
$$
\|A + B\| = \sup_{\|x\|=1} \| (A + B) x \| \leq \sup_{\|x\|=1} \left( \|A x\| + \|B x\| \right)
$$
by the triangle inequality for vector norms.

**Explanation:**  
Since $\|A x + B x\| \leq \|A x\| + \|B x\|$, the supremum over all unit vectors yields:  
$$
\|A + B\| \leq \sup_{\|x\|=1} \|A x\| + \sup_{\|x\|=1} \|B x\| = \|A\| + \|B\|
$$

**Final:**  
$\boxed{\|A + B\| \leq \|A\| + \|B\|}$.

---

### (d) Homogeneity: $\|\lambda A\| = |\lambda| \|A\|$, for scalar $\lambda \in \mathbb{R}$.

**Step 1:**  
By definition,  
$$
\|\lambda A\| = \sup_{\|x\|=1} \| (\lambda A) x \| = \sup_{\|x\|=1} |\lambda| \|A x\| = |\lambda| \sup_{\|x\|=1} \|A x\| = |\lambda| \|A\|
$$

**Explanation:**  
Scalar multiplication factors out of the norm.

**Final:**  
$\boxed{\|\lambda A\| = |\lambda| \|A\|}$.

---

### (e) Sub-multiplicativity: $\|A z\| \leq \|A\| \|z\|$.

**Step 1:**  
Since $\|A\| = \sup_{\|x\|=1} \|A x\|$, for any $z \in \mathbb{R}^n$, write $z$ as $z = \|z\| \cdot (z / \|z\|)$. Then,  
$$
\|A z\| = \|z\| \cdot \left\| A \frac{z}{\|z\|} \right\| \leq \|z\| \cdot \|A\| \quad \text{since} \quad \left\| A \frac{z}{\|z\|} \right\| \leq \|A\|
$$

**Explanation:**  
This shows the operator norm bounds the transformation of any vector $z$.

**Final:**  
$\boxed{\|A z\| \leq \|A\| \|z\|}$.

---

### (f) Submultiplicativity for matrices: $\|A B\| \leq \|A\| \|B\|$.

**Step 1:**  
For $A, B \in M_{m \times n}(\mathbb{R})$,  
$$
\|A B\| = \sup_{\|x\|=1} \|A B x\| \leq \sup_{\|x\|=1} \|A\| \|B x\| \leq \|A\| \sup_{\|x\|=1} \|B x\| = \|A\| \|B\|
$$

**Explanation:**  
The operator norm satisfies this submultiplicative property.

**Final:**  
$\boxed{\|A B\| \leq \|A\| \|B\|}$.

---

### (g) Spectral property: $\|A\|$ is the square root of the largest eigenvalue of $A^T A$.

**Step 1:**  
Recall that for real matrices, the operator norm can be written as:  
$$
\|A\| = \sup_{\|x\|=1} \|A x\| = \sqrt{\sup_{\|x\|=1} x^T A^T A x}
$$
Since $A^T A$ is symmetric and positive semi-definite, by the Rayleigh quotient,  
$$
\sup_{\|x\|=1} x^T A^T A x = \lambda_{\max}(A^T A)
$$
the largest eigenvalue of $A^T A$.

**Explanation:**  
Thus,  
$$
\|A\| = \sqrt{\lambda_{\max}(A^T A)}
$$

**Final:**  
$\boxed{\|A\| = \sqrt{\text{largest eigenvalue of } A^T A}}$.

---

**Summary of answers:**

- (a) $\|A\|$ is finite for all $A$.
- (b) $\|A\|=0 \iff A=0$.
- (c) $\|A + B\| \leq \|A\| + \|B\|$.
- (d) $\|\lambda A\| = |\lambda| \|A\|$.
- (e) $\|A z\| \leq \|A\| \|z\|$.
- (f) $\|A B\| \leq \|A\| \|B\|$.
- (g) $\|A\|$ equals the square root of the largest eigenvalue of $A^T A$.

VIPSolutions · Answer

Let's work through each part step-by-step:

**(a)** *Show that* $\|A\|$ *is finite for* $A \in M_{m \times n}(\mathbb{R})$:

- **Step 1:**  
By definition,  
$$
\|A\| = \sup_{\|x\|=1} \|A x\|
$$  
Since $A$ is a finite matrix, $\|A x\|$ is a continuous function of $x$, and the unit sphere $\{ x : \|x\|=1 \}$ is compact. The supremum of a continuous function over a compact set is finite, so $\|A\|$ is finite.

- **Explanation:**  
The operator norm measures the maximum stretching of unit vectors, which must be bounded for finite matrices.

**Final answer:** $\boxed{\|A\| < \infty}$.

---

**(b)** *Show that* $\|A\|=0$ *if and only if* $A=0$.

- **Step 1:**  
If $A=0$, then for all $x$, $A x=0$, so $\|A x\|=0$. Hence, $\|A\|=0$.

- **Step 2:**  
Conversely, if $\|A\|=0$, then for all $x$ with $\|x\|=1$, $\|A x\|=0$. By homogeneity, for any $x$,  
$$
A x = \|x\| A \left(\frac{x}{\|x\|}\right) = 0
$$
so $A x=0$ for all $x$, implying $A=0$.

**Final answer:** $\boxed{\|A\|=0 \iff A=0}$.

---

**(c)** *Prove the triangle inequality:* $\|A + B\| \leq \|A\| + \|B\|$.

- **Step 1:**  
For any $x$ with $\|x\|=1$,  
$$
\|(A+B) x\| \leq \|A x\| + \|B x\| \leq \|A\|\|x\| + \|B\|\|x\| = \|A\| + \|B\|
$$
since $\|A x\| \leq \|A\|\|x\|$.

- **Step 2:**  
Taking the supremum over all $\|x\|=1$,  
$$
\|A+B\| = \sup_{\|x\|=1} \|(A+B) x\| \leq \|A\| + \|B\|
$$

**Final answer:** $\boxed{\|A+B\| \leq \|A\| + \|B\|}$.

---

**(d)** *Show that* $\|\lambda A\| = |\lambda|\|A\|$.

- **Step 1:**  
By definition,  
$$
\|\lambda A\| = \sup_{\|x\|=1} \|(\lambda A) x\| = |\lambda| \sup_{\|x\|=1} \|A x\| = |\lambda| \|A\|
$$

**Final answer:** $\boxed{\|\lambda A\|= |\lambda| \|A\|}$.

---

**(e)** *Show that* $\|A z\| \leq \|A\| \|z\|$.

- **Step 1:**  
For any $z \neq 0$, write $z = \|z\| \frac{z}{\|z\|}$. Then,  
$$
\|A z\| = \|z\| \left\| A \frac{z}{\|z\|} \right\| \leq \|z\| \|A\| \quad \text{since} \quad \left\|A \frac{z}{\|z\|}\right\| \leq \|A\|
$$

**Final answer:** $\boxed{\|A z\| \leq \|A\| \|z\|}$.

---

**(f)** *Show that* $\|A B\| \leq \|A\| \|B\|$.

- **Step 1:**  
For matrices $A, B$,  
$$
\|A B\| = \sup_{\|x\|=1} \|A B x\| \leq \sup_{\|x\|=1} \|A\| \|B x\| \leq \|A\| \sup_{\|x\|=1} \|B x\| = \|A\| \|B\|
$$

**Final answer:** $\boxed{\|A B\| \leq \|A\| \|B\|}$.

---

**(g)** *Show that* $\|A\|$ *equals the square root of the largest eigenvalue of* $A^T A$.

- **Step 1:**  
Since $A^T A$ is symmetric positive semi-definite, the spectral theorem applies.  
$$
\|A\| = \sup_{\|x\|=1} \|A x\| = \sup_{\|x\|=1} \sqrt{x^T A^T A x} = \sqrt{\sup_{\|x\|=1} x^T A^T A x}
$$
By the Rayleigh quotient,  
$$
\sup_{\|x\|=1} x^T A^T A x = \lambda_{\max}(A^T A)
$$
the largest eigenvalue of $A^T A$.

- **Final answer:** $\boxed{\|A\| = \sqrt{\lambda_{\max}(A^T A)}}$.

---

**Summary:**

1. $\|A\|$ is finite.
2. $\|A\|=0 \iff A=0$.
3. $\|A + B\| \leq \|A\| + \|B\|$.
4. $\|\lambda A\| = |\lambda| \|A\|$.
5. $\|A z\| \leq \|A\| \|z\|$.
6. $\|A B\| \leq \|A\| \|B\|$.
7. $\|A\| = \sqrt{\lambda_{\max}(A^T A)}$.

That completes all parts!

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