1
$\begingroup$

I am reading Matrix Analysis and Applied Linear Algebra by Meyer and the following statement and proof are given enter image description here

What I am having trouble understanding is how the author is showing that $\beta_{\mathcal L}$ is a linearly independent set. Essentially, he states that the set of the transformations $\mathbf B_{ji}$ that act on an arbitrary but fixed element of $\beta$ is linearly independent. How does it follow from this premise that the entire set $\beta_{\mathcal L}$ is linearly independent?

$\endgroup$
1
$\begingroup$

The author is proving that for $\sum_{j,i}\eta_{j,i}\mathbf B_{j,i}=0$ to hold, all $nm$ coefficients $\eta_{j,i}$ have to be zero, since that is the definition of linear independence. This is shown in groups of $m$ coefficients at a time, namely for fixed $k\in\{1,\ldots,n\}$ it is shown that $\eta_{k,1},\ldots,\eta_{k,m}$ must all be zero. The latter is achieved by applying both sides of the given identity to $\mathbf u_k$: the right hand side is zero of course, and the left hand side gives a linear combination of $\mathbf v_1,\ldots,\mathbf v_m$ with $\eta_{k,1},\ldots,\eta_{k,m}$ as coefficients; by the linear independence of $\mathbf v_1,\ldots,\mathbf v_m$ this is only possible if all those $m$ coefficients are zero, as desired.

$\endgroup$
1
$\begingroup$

To say that vectors $\def\\#1{{\bf#1}}\\z_1,\ldots,\\z_n$ are independent (in your situation or any other) means that if $$\lambda_1\\z_1+\cdots+\lambda_n\\z_n=\\0$$ then $$\lambda_1=0\ ,\ldots,\ \lambda_n=0\ .$$ This is exactly what the author does. He starts with $$\sum_{j,i}\eta_{ji}\\B_{ji}=\\0$$ and proves that every $\eta_{ji}$ is zero. The introduction of the $\\u_k$ is just a convenient way to prove this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.