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Extending a linearly independent set

Let \(V\) be a vector space. Suppose that \(v_1,\ldots, v_m\) are linearly independent vectors in \(V\). If \(u \in V\) is not in the span of \(\{v_1,\ldots, v_m\}\), then \(\{u, v_1,\ldots, v_m\}\) is linearly independent.

To see this, suppose that \(\{u, v_1,\ldots, v_m\}\) is not a linearly independent set. Hence, there exist scalars \(\mu, \lambda_1,\ldots, \lambda_m\) not all equal to zero such that \[\mu u + \lambda_1 v_1 + \cdots + \lambda_m v_m = 0_V.\] Note that \(\mu\) cannot be \(0\) or else we would have a linear combination of \(v_1,\ldots, v_m\) with scalars not all equal to \(0\) that give the zero vector, contradicting that \(v_1,\ldots, v_m\) are linearly independent. Hence, we can rewrite the equation as \[u = \frac{-\lambda_1}{\mu} v_1 + \cdots +\frac{-\lambda_m}{\mu} v_m.\] This contradicts that \(u\) is not in the span of \(\{v_1, \ldots, v_m\}\). Therefore, \(\{u, v_1,\ldots, v_m\}\) must be linearly independent.

The result above shows that one can obtain a basis for \(V\) by starting with a linearly independent set of vectors and repeatedly adding a vector not in the span of the vectors to the set until it spans \(V\). The resulting set will be a basis for \(V\) since it is linearly independent and spans \(V\).

This also means that if \(B\) is a linearly independent set of vectors with \(|B| = \dim(V)\), then \(B\) must span \(V\) because otherwise, \(B\) can be extended to a basis for \(V\) with more than \(\dim(V)\) vectors, contradicting that all bases for \(V\) have the same number of vectors.

Dimension of a subspace

Let \(W\) be a subspace of \(V\). An immediate consequence of the above is that \(\dim(W) \leq \dim(V)\). To see this, let \(w_1,\ldots,w_m\) be a basis of \(W\) where \(m = \dim(W)\). As \(W\) is a subspace of \(V\), \(\{w_1,\ldots,w_m\}\) is a linearly independent set in \(V\) and its span, which is simply \(W\), is contained in \(V\). Extend this set to \(\{w_1,\ldots,w_m,u_1,\ldots, u_k\}\) so that it gives a basis for \(V\). Then \[m + k = \dim(V).\] As \(k \geq 0\), we get \(m \leq \dim(V)\), with strict inequality if and only if \(W\) is a proper subspace of \(V\).

Example

Suppose that we are asked to extend \(U = \left\{\begin{bmatrix} 1\\1 \\0 \end{bmatrix}, \begin{bmatrix} 1\\0 \\-1 \end{bmatrix}\right\}\) to a basis of \(\mathbb{R}^3\). Since the dimension of \(\mathbb{R}^3\) is three and \(U\) already contains two linearly independent vectors, all we need to do is to find a vector in \(\mathbb{R}^3\) that is not in the span of \(U\). The question is, how do we find such a vector when there are infinitely many vectors in \(\mathbb{R}^3\) to choose from?

We will do something clever by making the following observation: As \(\mathbb{R}^3\) is spanned by \(\left\{\begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix} 0\\1\\0\end{bmatrix}, \begin{bmatrix} 0\\0\\1\end{bmatrix}\right\}\), there must be at least one vector in the set that cannot be written as a linear combination of vectors in \(U\). Otherwise the span of \(U\) will contain \(\mathbb{R}^3\), which is impossible because there are only two vectors in \(U\).

In other words, we are trying to solve \(Ax = b\) where the columns of \(A\) are the vectors in \(U\) and \(x = \begin{bmatrix} x_1\\x_2\end{bmatrix}\) for each choice of \(b\) from \(\left\{\begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix} 0\\1\\0\end{bmatrix}, \begin{bmatrix} 0\\0\\1\end{bmatrix}\right\}\). Since all three systems have the same coefficient matrix, we can simply row reduce the extended augmented matrix: \[\left[ \begin{array}{rr|rrr} 1 & 1 & 1 & 0 & 0\\ 1 & 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0 & 1 \end{array} \right].\] Interchanging rows 1 and 2, we obtain \[\left[ \begin{array}{rr|rrr} 1 & 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 0 & 0\\ 0 & -1 & 0 & 0 & 1 \end{array} \right].\] Performing \(R_2 \leftarrow R_2 - R_1\) gives \[\left[ \begin{array}{rr|rrr} 1 & 0 & 0 & 1 & 0\\ 0 & 1 & 1 & -1 & 0\\ 0 & -1 & 0 & 0 & 1 \end{array} \right].\] Performing \(R_3 \leftarrow R_3 + R_2\) gives \[\left[ \begin{array}{rr|rrr} 1 & 0 & 0 & 1 & 0\\ 0 & 1 & 1 & -1 & 0\\ 0 & 0 & 1 & -1 & 1 \end{array} \right].\] From the bottom row, we see that none of the three vectors can be written as a linear combination of vectors in \(U\). That means we could add any of them to \(U\). The resulting set will have three linearly independent vectors and will therefore span \(\mathbb{R}^3\).

Quick Quiz

Exercise

  1. Let \(P_2\) denote the vector space of polynomials in \(x\) with real coefficients having degree at most \(2\). Extend \(\{x-1, x^2-x + 2\}\) to a basis of \(P_2\).  

  2. What is the largest possible dimension of a proper subspace of the vector space of \(2 \times 3\) matrices with real entries?  

  3. Let \(B = \{v_1,\ldots, v_n\}\) be a basis of \(\mathbb{R}^n\). \(B\) is said to be an orthogonal basis if the dot product of \(v_i\) and \(v_j\) is \(0\) for all distinct \(i, j \in \{1,\ldots, n\}\). For example, the basis \(\left\{\begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix} 0\\1\\0\end{bmatrix}, \begin{bmatrix} 0\\0\\1\end{bmatrix}\right\}\) is an orthogonal basis of \(\mathbb{R}^3\).

    1. Extend \(\left\{\begin{bmatrix} 1\\-2\end{bmatrix}\right\}\) to an orthogonal basis of \(\mathbb{R}^2\).  

    2. Extend \(\left\{\begin{bmatrix} 1\\0\\ -1 \end{bmatrix}, \begin{bmatrix} 1\\2\\ 1 \end{bmatrix}\right\}\) to an orthogonal basis of \(\mathbb{R}^3\).