Understanding Basis for Solving Linear Algebra Problems

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ElliottG
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Hey guys

There are so many of these damn "Find a basis" questions and I can't get any of them because we never directly learned how...or she never showed us in class...my final exam is tomorrow.

Here are some examples of questions:

http://184.154.165.18/~devilthe/uploads/1323453294.png

http://184.154.165.18/~devilthe/uploads/1323430492.png

Part D[/color]

I have zero idea how they solve these...

I know that a basis is a linearly independent spanning set...but how to solve these questions? No idea.

Can there be more than one basis for a question?

Thanks,
Elliott
 
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A basis is a set of vectors (or matrices in this problem) that is linearly independent and spans the (sub)space. A vector (sub)space can have many bases.

For d, start by playing with the equation that defines the subspace V, using an arbitrary matrix X, where
X = \begin{bmatrix}a & b \\ c & d \end{bmatrix}
 
Any polynomial, of degree 2, is of the form p(x)= ax^2+ bx+ c. p'(x)= 2ax+ b Requiring that p'(1)= 0 means that 2a+ b= 0 so b= -2a. That is, for any p in this set, p(x)= ax^2- 2ax+ c= a(x^2- 2)+ c(1). Now, what is a basis for that set?

Any 2 by 2 matrix is of the form
\begin{bmatrix}a & b \\ c & d\end{bmatrix}

We require that
\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix}= \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}

Multiply those, set corresponding term equal, and see what you get.
 
Thanks guys.

Anyway I just did my linear exam today so hopefully I never have to see linear ever again!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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