# Help with linear algebra question about basis for a vector space.

Let $\{u_1, u_2, u_3\}$ be a basis for a vector space $V$. Then if we define $v_1 = u_1 + 2u_3$, $v_2 = u_1 + 2u_2 + 3u_3$, $v_3 = u_2 - u_3$, show that $\{v_1, v_2, v_3\}$ forms a basis for $V$.

Attempt:

I honestly have no idea where to start. All I know is that a basis for a vector space is the minimum amount of vectors that represent a set of vectors and that the vectors are linearly independent.

• Well you know it has enough cardinality. Just show it is linearly independent by trying to solve the linear equation that would show dependency and fail. – Jacob Wakem Nov 3 '16 at 5:31

Since $(u_1,u_2,u_3)$ form a basis of the vector space $V$ we know dim(V)=3. This means that any linearly independent set of 3 vectors would be a basis of the vector space. Show that the defined $(v_1,v_2,v_3)$ are linearly independent. (Remember that we already know $(u_1,u_2,u_3)$ are linearly independent).

There are multiple ways of answering this, but for a lazy person with access to a computer or calculator, I'd first rewrite this in terms of matrices. For example, for $U = [u_1,u_2,u_3]$ and $V = [v_1,v_2,v_3]$, then $V=UA$ where

$A = \left[\begin{matrix} 1 &1&0\\0&2&1\\2&3&-1\end{matrix}\right]$

Since $v_i$'s is written as a linear combination of $u_i$'s, then the subspace spanned by $v_i$'s is a subset of the subspace spanned by $u_i$.

Now, if $A$ is full rank (which can be verified in a number of ways, using rank(A), SVD or rref by hand, or any other method of choice), then $A$ is invertible, so $U = VA^{-1}$. So, the subspace spanned by $u_i$'s is also a subset of the subspace spanned by $v_i$'s.

Therefore, the two span the same subspace. Since 3 $u_i$'s are all independent, they must span a vector space of dim 3, which means 3 $v_i$'s spanning the same subspace must be linearly independent. So, $v_i$'s form a basis.

• What do you mean by "A is full rank" ? Rank is the number of linearly independent rows right ? – YDDOR Nov 3 '16 at 5:52
• I think full rank usually refers to full column rank but in this case it doesn't matter, since for square matrices row rank = column rank. Additionally for square matrices, full rank implies the matrix is invertible, which is the quality you need. – Y. S. Nov 3 '16 at 7:34

Since $\{u_1,u_2,u_3\}$ is a basis for $V$, it follows that $V$ has dimension $\color{red}{3}$.

You are given a set of vectors $\{v_1,v_2,v_3\}$. Clearly, this set contains $\color{red}{3}$ vectors in $V$. Hence, in order to deduce that it is a basis for $V$, all you have to do is prove that $v_1,v_2,v_3$ are linearly independent.

You can do so using the definition of linear independency as follows. Suppose that there exists three scalars $\gamma_1,\gamma_2,\gamma_3$ such that $$\gamma_1\color{blue}{v_1}+\gamma_2\color{green}{v_2}+\gamma_3\color{orange}{v_3}=0.$$ (Note that the $0$ on the RHS of the equality is the zero vector.) The goal is to show that $\gamma_i=0$ (here $0$ is the zero scalar) for all $i$. We can rewrite the above equality as $$\gamma_1\color{blue}{(u_1+2u_3)}+\gamma_2(\color{green}{u_1+2u_2+3u_3})+\gamma_3(\color{orange}{u_2-u_3})=0,$$ or $$(\gamma_1+\gamma_2)u_1+(2\gamma_2+\gamma_3)u_2+(2\gamma_1+3\gamma_2-\gamma_3)u_3=0.$$ Now, you know that $u_1,u_2,u_3$ are linearly independent, so the coefficients must all equal zero: \begin{align} \gamma_1+\gamma_2=0,\\ 2\gamma_2+\gamma_3=0,\\ 2\gamma_1+3\gamma_2-\gamma_3=0. \end{align} This is a linear system that is readily solved for $\gamma_1,\gamma_2,\gamma_3$. Indeed, the first equation gives $\gamma_2=-\gamma_1$, the second one then gives $\gamma_3=2\gamma_1$ and the third one leads to $-3\gamma_1=0$. All in all, we do must have $\gamma_1=\gamma_2=\gamma_3=0$, which was to be proved.