Lecture 33

[[lecture-data]]

2024-11-18

Exam december 4:

• Chapter 5 and 6
• Norms, Gersgorin discs, today we’ll talk about positive matrices
• cohesive subject material
• We’ll go through what we should know

6. Chapter 6

Recall from last time that non-interior eigenvalues of ALL gersgorin discs of irreducible matrices are on the boundary of ALL discs

Theorem

Let $A\in M_n$ be irreducible. If $\lambda \in \sigma (A)$ is not interior of any Gersgorin Disc, then $\lambda$ is on the boundary of ALL Gersgorin discs.

Last time, we showed that irreducible diagonally dominant (with strictness in one row) matrices are invertible, which was a relaxation (or substitution) of the theorem showing that strictly diagonally dominant matrices are invertible.

Corrolary

Suppose $A\in M_n$ is irreducible. Suppose there exists some index $p$ such that ${\sum }_j|a_{pj}|<||A|{|}_{\infty ,\infty }$ (ie there is some row sum that is strictly less than the maximum row sum). Then $\rho (A)<||A|{|}_{\infty ,\infty }$

$\exists \lambda \in \sigma (A)$ such that $|\lambda |=||A|{|}_{\infty ,\infty }$. Then for all $i$, we have by the triangle inequality $|\lambda -a_{ii}|\ge |\lambda |-|a_{ii}|$ and

BWOC, suppose

|\lambda-a_{ii}| &\geq |\lambda| - |a_{i i}| \\ &=\lvert \lvert A \rvert \rvert _{\infty, \infty} - |a_{i i}| \\ &\geq \left( \sum_{j} |a_{ij}| \right) - |a_{ii}| \\ &= \sum_{j \neq i} |a_{i j}| \\ \end{aligned}$$ Thus for all $i$, $\lambda$ is [[Concept Wiki/eigenvalue interior of gersgorin disc\|not interior]] to the associated [[Concept Wiki/Gersgorin Disc]]. So by irreducibility of $A$ and since [[Concept Wiki/non-interior eigenvalues of ALL gersgorin discs of irreducible matrices are on the boundary of ALL discs]], we must have that $\lambda$ is on the boundary of ALL Gersgorin discs of $A$. In particular, equality holds in all the inequalities of $|\lambda-a_{ii}| \geq |\lambda| - |a_{i i}|$. But then there is no $p$th row that is less than the maximum row sum $\implies \impliedby$. Thus there must not exist such a $\lambda$.

(see irreducible matrices with row sums not all equal have spectral radius less than the maximum row sum)

8. Chapter 8

Nonnegative and positive matrices (see Chapter 8)

Notation

• We will still use the notation $A,B\in M_n$, but for this chapter they are ALWAYS real.
• We will look at the spectrum of these matrices, so their eigenvalues and eigenvectors may be complex.
• "$A>0$" means entrywise, for all $i,j$ we have $a_{ij}>0$. Same for $\ge$
• "$A>B$" means entrywise for all $i,j$ we have $a_{ij}>b_{ij}$. Same for $\ge$
• "$|A|$" means I want to take the entrywise absolute value of $A$. So this is a matrix where $\forall i,j$, $|A{|}_{ij}=|A_{ij}|$. And this can be used for complex-valued matrices as well.

Nonnegative and Positive Matrices

"$A>0$" means entrywise, for all $i,j$ we have $a_{ij}>0$. Same for $\ge$. If $A>0$, then $A$ is a positive matrix.

"$A>B$" means entrywise for all $i,j$ we have $a_{ij}>b_{ij}$.

"$|A|$" means I want to take the entrywise absolute value of $A$. So this is a matrix where $\forall i,j$, $|A{|}_{ij}=|A_{ij}|$. And this can be used for complex-valued matrices as well.

(see positive matrix)

Observation

Observe that $|AB|\le |A|\cdot |B|$ and also $|Ax|\le |A|\cdot |x|$. This is just an immediate application of the triangle inequality component-wise for these matrices.

Further, for any positive integer $k$, we have $|A^k|\le |A{|}^k$ by immediate application of the above.

Finally, if $0\le A\le B$, then $0\le A^k\le B^k$

(see notes on positive matrices)

Proposition

Suppose $A\in M_n$ and $A\ge 0$ ie nonnegative matrix. If for all $i$ we have ${\sum }_j{a}_{ij}=\alpha$ (for some $\alpha \in \mathbb{R}$) ie all the row sums are the same, then $\alpha =\rho (A)=||A|{|}_{\infty ,\infty }$

$e$ be the vector of all 1s. Then $Ae=\alpha e$. So $\alpha \in \sigma (A)$ with eigenvector $e$ and $\alpha \le \rho (A)\le ||A|{|}_{\infty ,\infty }=\alpha$, where the second inequality comes from matrix norms are bounded below by the spectral radius.

Let

(see nonnegative matrices with equal row sums have row sum equal to the spectral radius)

Theorem

Let $A,B\in M_n$ such that $|A|<B$. (Then $B$ must be a nonnegative matrix). Then $\rho (A)\le \rho (|A|)\le \rho (B)$.

$k$, we have $|A^k|\le |A{|}^k\le B^k$ (see notes on positive matrices). Since the frobenius norm is absolute (monotone), we have $|||A^k||{|}_F\le |||A{|}^k|{|}_F\le ||B^k|{|}_F$ And this yields $|||A^k||{|}_F^{1/k}\le |||A{|}^k|{|}_F^{1/k}\le ||B^k|{|}_F^{1/k}$ Now, let $k\to \infty$. For each $k$, the above relationship holds. Thus, this relationship will hold at the limit. And since the limit of the powered norm is the spectral radius, we have that $\rho (A)\le \rho (|A|)\le \rho (B)$

For all indices

(see matrices that dominate nonnegative matrices have dominant spectral radius)

Corrolary

Let $A\in M_n$ such that $A$ is nonnegative matrix. Then ${\min }_i{\sum }_j{a}_{ij}\le \rho (A)$

$\rho (A)\le {\max }_i\sum a_{ij}=||A|{|}_{\infty ,\infty }$ since matrix norms are bounded below by the spectral radius.

By way of comparison, we have

${\min }_i{\sum }_j{a}_{ij}=0$ then the result is trivial.

If

Otherwise, obtain $\tilde{A}$ from $A$ where for each $k=1,2,\dots ,n$, multiply the $k$th row of $A$ by $\frac{{\min }_i{\sum }_j{a}_{ij}}{{\sum }_j{a}_{kj}}\le 1$. Note that each row sum of $\tilde{A}$ is equal to the minimum!

Further, since each row sum of $\tilde{A}$ is ${\min }_i{\sum }_j{a}_{ij}$ , we have that ${\min }_i{\sum }_j{a}_{ij}=\rho (\tilde{A})$ since nonnegative matrices with equal row sums have row sum equal to the spectral radius.

Finally, since each scaling factor is $\le 1$, we have that $0\le \tilde{A}\le A$. And since matrices that dominate nonnegative matrices have dominant spectral radius, we get that ${\min }_i{\sum }_j{a}_{ij}=\rho (\tilde{A})\le \rho (A)$

(see the min row sum is a lower bound for the spectral radius of a nonnegative matrix)

Note

Since the min row sum is a lower bound for the spectral radius of a nonnegative matrix, if $A>0$ ie $A$ is a positive matrix, then $\rho (A)>0$. ( we can also see this since matrices are nilpotent iff all eigenvalues 0)

(see positive matrices have positive spectral radius)

Note

And if $A\ge 0$ and irreducible, then $\rho (A)>0$.

$A$ is irreducible, then it has no rows of all zero. Thus each row sum must be strictly positive. And since the min row sum is a lower bound for the spectral radius of a nonnegative matrix, this means the lower bound for the spectral radius is positive.

If

(see nonnegative, irreducible matrices have positive spectral radius)

Theorem

Let $A\in M_n,x\in {\mathbb{C}}^n$ with $A\ge 0,x>0$. If there exists $\alpha \ge 0$ such that

• $Ax>\alpha x$, then $\rho (A)>\alpha$.

• $Ax\ge \alpha x$, then $\rho (A)\ge \alpha$.

• $Ax<\alpha x$, then $\rho (A)<\alpha$.

• $Ax\le \alpha x$, then $\rho (A)\le \alpha$.

Proof

To come next time.

(see )