# Basis (linear algebra)

Let's have vectors $v_1,v_2,v_3,v_4\in \mathbb R^4$. Prove that these vectors form basis (1), determine if they are orthogonal and/or orthonormal (2). Then make a transition matrix $T_{\epsilon\alpha}$ from basis $\alpha$ to standard basis and use it to get coordinates of vector $w=(2,3,5,1)_\alpha$ in standard basis (3). $$v_1=(1,1-1,1),v_2=(1,-1-1,-1),v_3=(0,1,0,-1),v_4=(1,0,1,0)$$ (1) Prove that these vectors form a basis.

I did following: $$\left( \begin{array}{cccc} 1 & 1 & -1 &1 \\ 1 & -1 & -1 &-1 \\ 0 & 1 & 0 &-1 \\ 1 & 0 & 1 & 0 \end{array} \right)=A$$ $$\left| \begin{array}{cccc} 1 & 1 & -1 &1 \\ 1 & -1 & -1 &-1 \\ 0 & 1 & 0 &-1 \\ 1 & 0 & 1 & 0 \end{array} \right|=-8 \$$ $|A| \neq 0 \implies$ Vectors $v_1,v_2,v_3,v_4$ are linearly independent and form basis.

(2) Determine if they are orthogonal and/or orthonormal.

I did following: $$v_1\cdot v_2=(1-1+1-1)=0$$ $$v_1\cdot v_3=(0+1+0-1)=0$$ $$v_1\cdot v_4=(1+0-1+0)=0$$ $$v_2\cdot v_3=(0-1+0+1)=0$$ $$v_3\cdot v_4=(0+0+0+0)=0$$ $\implies$ Basis is orthogonal. $$|v_1|=\sqrt {1+1+1+1}=\sqrt4=2$$ $$|v_2|=\sqrt {1+1+1+1}=\sqrt4=2$$ $$|v_3|=\sqrt {0+1+0+1}=\sqrt2$$ $$|v_4|=\sqrt {1+0+1+0}=\sqrt2$$ $\implies$ Basis is not orthonormal.

(3) Make a transition matrix $T_{\epsilon\alpha}$ from basis $\alpha$ to standard basis and use it to get coordinates of vector $w=(2,3,5,1)_\alpha$ in standard basis.

$$T_{\epsilon\alpha}=A^{-1}=\left( \begin{array}{cccc} \frac14 & \frac14 & 0 & \frac12 \\ \frac14 & -\frac14 & \frac12 &0 \\ -\frac14 & -\frac14 & 0 & \frac12 \\ \frac14 & -\frac14 & -\frac12 & 0 \end{array} \right)$$

$$u=A^{-1} w$$ Where $u$ is in standard basis?

Is (1) and (2) correct and can you help me out with (3) ?

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It's important to remember that a vector $w$ written in terms of basis $\alpha = \{v_1, v_2, v_3, v_4\}$ has the form:

$$( a, b, c, d)_{\alpha} = av_1 + bv_2 + cv_3 + dv_4$$

It's also important to remember that when your vectors $v_i$ are written in terms of coordinates, that these are coordinates with respect to the standard basis. For example,

$$( 1, 0, 0, 0)_{\alpha} = v_1 = (1, 1, -1, 1)_{\epsilon}$$

Therefore, the matrix $T_{\epsilon \alpha}$ should have the property that:

$$T_{\epsilon \alpha} (a,b,c,d) = a( 1, 1, -1, 1)+ b( 1, -1, -1,-1) + c( 0, 1, 0, -1)+ d( 1, 0, 1, 0)$$

Thus, $T_{\epsilon \alpha} = A$, the matrix you've written above, whose rows are the standard-basis representations of the vectors $v_i$ in the given order.

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So $T_{\alpha\epsilon}=A^{-1}$ and $u=w\cdot T_{\alpha\epsilon}$ where $u$ is $w$ in standard basis? – MartinB Dec 5 '12 at 13:32
@MartinB, $u = w\cdot T_{\alpha \epsilon}$ would express a vector $w$ which is given in standard $\epsilon$ basis as a vector $u$ in basis $\alpha$. – Shaun Ault Dec 5 '12 at 13:42
@MartinB: Also, your (1) and (2) calculations are fine... except you forgot to check that $v_2 \cdot v_4 = 0$. – Shaun Ault Dec 5 '12 at 13:43
So $u = w\cdot T_{\epsilon\alpha}$, where $u$ is in standard basis. Thanks a lot for help. – MartinB Dec 5 '12 at 13:45
@ShaunAult Is the vector $u =(1,-7,2,7)$ ? MartinB is sitting next to me :) – Noturab Dec 5 '12 at 14:28

Let $A$ be the matrix that transforms the standard basis to $\alpha$. Then the columns of $A$ are your vectors $v_1,v_2,v_3,v_4$. Also, for each standard basis element $e_i$, we have $Ae_i=v_i$. Thus $e_i=A^{-1}v_i$. Therefore you need to calculate $A^{-1}$.

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I've edited my question, is it correct now? And what about (1) and (2)? – MartinB Dec 5 '12 at 13:19