Cauchy-Schwarz inequality and the frobenius norm

I’m reading the Linear Algebra book and I’m finding it difficult to understand the relationship between the cauchy schwarz inequality and the frobenius norm in chapter 6.

I don’t get why the norm of the matrix vector multiplication is taken to be the sum of the absolute values of it’s individual elements, given that the matrix-vector multiplication is itself a vector and norm of a vector is the square root of dot product with itself.


Hi, sorry if that wasn’t explained clearly enough in the book. It’s not the magnitudes of the individual matrix elements; it’s the vector dot product between each row i of the matrix with the vector v. So, a_i is the i^th row of matrix A.

Perhaps it’s confusing because you’re used to seeing the dot product written as a^T v, so the first vector transposed. That assumes the vectors are both columns; here, a_i is already a row vector, so the dot product is simply av.


Thanks for the reply.

There is obviously something that I’m not getting.
I was expecting ||Av|| to be equal to a sum of squares ie. (a_1v)^2 + (a_2v)^2 + …
since it is the dot product rather than a sum of absolute values.

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Oh damn, now I see what you mean. Thank you for your persistence and my apologies for not looking at this carefully enough in your first question.

Indeed, you are correct, and there’s a typo in the book. In fact, all the terms in equations 6.28 and 6.29 should be squared. I have no good excuse for that; I think I just missed it. (You could also square the terms in 6.27, but that doesn’t matter because squaring is a monotonic function so the inequality still holds.)

I’ll fix that in the book so it’s correct in the next version. Thanks again for bringing that to my attention!

Hi, thanks for the clarification. I appreciate your help so much!