Solution Space of a Homogeneous System of Linear Equations

Let $A$ be an $m\times n$ matrix.

<aside> 💡 The subspace $W=\{ x\in \R^n |~~Ax=0 \}$ of $\R^n$ is called the solution space of the homogeneous system of linear equations $AX=0$ or the null space of $A$

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Note that the null space is a subspace of $\R^n$.

<aside> 💡 The dimension of the null space is called the nullity of $A$.

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To check whether $W$ is a subspace, Let $x,y \in W$ , then: $x+y\in W$ →(1) Let $\lambda \in \R$ , then : $A(\lambda x)=\lambda (Ax)=\lambda0=0$ , i.e. $\lambda x$ should also be in $W$.→(2)

Finding the Nullity and a Basis for the Null Space

We will use row reduction to also find the nullity and a basis for the null space of $A.$

Using Gaussian Elimination, Convert the augmented matrix $[A|0]$ for $AX=0$ to its row echelon form.

<aside> 💡 $nullity(A)$= number of independent variables in the row reduced echelon form of augmented matrix $[A|0]$.

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• Assign arbitrary value $t_i$ to the $i^{th}$ independent variable.
• Compute the value of each dependent variables in the terms of $t_i$s from the unique row it occurs in.
• Every solution is obtained by letting $t_i$s vary in $\R$.

<aside> 💡 The vectors obtained by substituting $t_i=1$ and $t_j=0 ~~\forall j \neq i$ as $i$ varies constitutes a basis of the null space of $A$.(i.e., the solution space of $Ax=0$).

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Notice that since the system is homogeneous, the augmented 0 vector remains unchanged during row reduction process. So augmentation is not required.

Example