# Finding values to make a set a basis

Question: For which values of $a\in \Bbb R$ is the set below a basis for $M_{2x2}(\Bbb R)$ as a vector space over $\Bbb R$? $${(a,2a,2,3a), (1,2,2a,3), (1,2a,a+1,a+2), (1,a+1,2,2a+1)}$$ (these vectors should be matrices but I do not know how to do them in MathJax)

My thinking is that in order to be a basis the set must be linearly independent. Thus the sum of the set of vectors and the set of scalar coefficients must equal zero where all scalar coefficients are equal to zero. If this is the case then any number could be $a$ without affecting the linear independence.

Is my thinking totally off? Can someone explain this problem if it is.

• Do you know determinants? Sep 7, 2016 at 23:29
• I do. Havnt done them in a while but I know how. Sep 7, 2016 at 23:32
• Compute the determinant of the system of vectors. Permuting rows, you can have a very nice pivot for a rather fast computation. Sep 7, 2016 at 23:35
• I will try that. Thank you! Sep 7, 2016 at 23:37
• Your statement that $a$ can be any number without affecting the linear independence is incorrect. A set of vectors is linearly independent if the only linear combination that results in a sum of zero occurs when all of the coefficients are zero. Thus, you are trying to find the specific values of $a$ such that this occurs. Sep 8, 2016 at 0:01

In order for the matrices presented to form a basis they need to be linearly independent. This means that for scalars $x_i$, the equation:

$$x_1 \begin{pmatrix} a & 2a \\ 2 & 3a \end{pmatrix} + x_2 \begin{pmatrix} 1 & 2 \\ 2a & 3 \end{pmatrix} + x_3 \begin{pmatrix} 1 & 2a \\ a+1 & a+2 \end{pmatrix} + x_4 \begin{pmatrix} 1 & a+1 \\ 2 & 2a+1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$

should have only the trivial solution. To solve this, we need to solve the system:

\begin{align} ax_1 + x_2 + x_3 + x_4 &= 0 \\ 2ax_1 + 2x_2 +2ax_3 +(a+1)x_4 &= 0 \\ 2x_1 + 2ax_2 + (a+1)x_3 + 2x_4 &= 0 \\ 3ax_1 +3x_2 +(a+2)x_3 + (2a+1)x_4 &= 0 \end{align}

Ensuring that this system has only the trivial solution amounts to making sure that $\det{A} \neq 0$ where:

$$A = \begin{pmatrix} a & 1 & 1 & 1 \\ 2a & 2 & 2a & a+1 \\ 2 & 2a & a+1 & 2 \\ 3a & 3 & a+2 &2a+1 \end{pmatrix}$$

After finding the determinant of $A$ you can find which $a$ values result in $\det{A} \neq 0$ It is precisely these values of $a$ which result in the set being a basis.

We can view elements of $M_{2 \times 2}(\mathbb{R})$ as column vectors under the standard isomorphism. Thus we have those 4 vectors are a basis if and only if the determinant of the matrix with those as the columns is nonzero.

We calculate

$$\det( \begin{bmatrix} a & 2a & 2 & 3a \\ 1 & 2 & 2a & 3 \\ 1 & 2a & a+1 & a+2 \\ 1 & a+1 & 2 & 2a + 1 \\ \end{bmatrix}) = 6 (1-a)^3 (1+a)$$

Hence its a basis when $a \neq \pm 1$