MHB Is the Condition Number of A'A Related to its Matrix Norm?

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Condition
AI Thread Summary
The discussion focuses on the relationship between the condition number of a matrix \( A \) and its transpose \( A^T \). It is established that the condition number \( k_2(A^TA) \) can be expressed as \( (k_2(A))^2 \). The participants explore the definitions of matrix norms and the implications of maximizing certain quadratic forms involving \( A \) and \( A^TA \). The conversation highlights the need for clarity in understanding how these norms and condition numbers relate to each other, particularly in the context of unit vectors. Ultimately, the discussion emphasizes the mathematical connections between the condition numbers and the properties of the matrices involved.
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

We have a matrix $A\in \mathbb{R}^{m\times n}$ which has the rank $n$. The condition number is defined as $\displaystyle{k(A)=\frac{\max_{\|x\|=1}\|Ax\|}{\min_{\|x\|=1}\|Ax\|}}$.

I want to show that $k_2(A^TA)=\left (k_2(A)\right )^2$. We have that $$k_2(A^TA)=\frac{\max_{\|x\|_2=1}\|(A^TA)x\|_2}{\min_{\|x\|_2=1}\|(A^TA)x\|_2}$$

It holds that $$\|(A^TA)x\|_2=\left ((A^TA)x, (A^TA)x\right )=\left (A^TAx\right )^T\left (A^TAx\right )=x^TA^TAA^TAx=\left (Ax\right )^T\left (AA^T\right )\left (Ax\right )$$
We also have that $$\|Ax\|_2=\left (Ax, Ax\right )=\left (Ax\right )^T\left (Ax\right )=x^TA^TAx=x^T\left (A^TA\right )x$$

How do we continue? (Wondering)
 
Mathematics news on Phys.org
Hey mathmari!

Shouldn't it be $\|Ax\|_2 = \sqrt{(Ax,Ax)}$ or $\|Ax\|_2^2 = (Ax,Ax)$? (Worried)

So we have:
$$\|Ax\|_2^2=x^T\left (A^TA\right )x$$

We also have:
$$\|(A^TA)x\|_2^2=x^TA^TAA^TAx=x^T(A^TA)^2x$$

Suppose $x$ is a vector of unit length for which $x^T\left (A^TA\right )x$ takes on its maximal value.
Then $x^T(A^TA)^2x$ will also take its maximal value, won't it? (Wondering)
 
Klaas van Aarsen said:
Shouldn't it be $\|Ax\|_2 = \sqrt{(Ax,Ax)}$ or $\|Ax\|_2^2 = (Ax,Ax)$? (Worried)

Ohh yes! (Blush)
Klaas van Aarsen said:
So we have:
$$\|Ax\|_2^2=x^T\left (A^TA\right )x$$

We also have:
$$\|(A^TA)x\|_2^2=x^TA^TAA^TAx=x^T(A^TA)^2x$$

Suppose $x$ is a vector of unit length for which $x^T\left (A^TA\right )x$ takes on its maximal value.
Then $x^T(A^TA)^2x$ will also take its maximal value, won't it? (Wondering)

Will $x^T(A^TA)^2x$ take also its maximal value because the term in the middle is the square of the previous one? (Wondering)
 
mathmari said:
Will $x^T(A^TA)^2x$ take also its maximal value because the term in the middle is the square of the previous one?

Well, to be fair, this is not immediately obvious. (Worried)

Let's take a slightly different angle.

According to the wiki page about matrix norms, a matrix $A$ has norm $\|A\|_2 =\sup\limits_{\|x\|=1} \|Ax\|_2 = \sqrt{\lambda_{\text{max}}(A^*A)}=\sigma_{\text{max}}(A)$, where $A^*$ is the conjugate transpose, which is just the transpose $A^T$ for a real matrix.

So $\|A^TA\|_2 = \sqrt{\lambda_{\text{max}}((A^TA)^T A^TA)}$, isn't it? (Wondering)
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top