By construction, \(V\) is the sum of its the given eigenspaces. It, therefore, only remains to show that these eigenspaces only have the 0-vector in common. In other words, if \(\Vect{a}_i\) is in \(\EigenSpc{\EigenVal{\lambda}_i}\), we need to show that
\[\Vect{a}_i + \cdots +\Vect{a}_k = \Vect{0} \quad\text{if and only if} \quad \Vect{a}_1=\cdots = \Vect{a}_k=\Vect{0}\]
Only the direction from left to right presents a challenge, and here we argue by induction on \(k\). For \(k=1\), the claim is a tautology. Now suppose the claim is true for \(1\leq k-1\). We need to infer its truth for \(k\). So suppose
\[\Vect{a}_1 + \cdots +\Vect{a}_k = \Vect{0}\]
Then
| \(A(\Vect{a}_1 + \cdots +\Vect{a}_k)\) | \(=\) | \(\Vect{0}\) |
| \(\EigenVal{\lambda}_1\cdot \Vect{a}_1 + \cdots + \EigenVal{\lambda}_k\cdot \Vect{a}_k\) | \(= \) | \(\Vect{0}\) |
Now we need the assumption that the eigenvalues \(\EigenVal{\lambda}_1, \dots , \EigenVal{\lambda}_k\) are distinct. Therefore, at least one of them is not 0. By rearranging indices, if necessary, we may achieve \(\EigenVal{\lambda}_k\neq 0\). Dividing the last equation by \(\EigenVal{\lambda}_k\), and recalling the original equation yields
| \(\Vect{a}_1 + \cdots + \Vect{a}_{k-1} + \Vect{a}_k\) | \(=\) | \(\Vect{0}\) |
| \(\tfrac{\EigenVal{\lambda}_1}{\EigenVal{\lambda}_k}\cdot \Vect{a}_1 + \cdots + \tfrac{\EigenVal{\lambda}_{k-1}}{\EigenVal{\lambda}_k}\cdot \Vect{a}_{k-1} + \Vect{a}_k\) | \(=\) | \(\Vect{0}\) |
Subtracting the first equation from the second yields
| \((\tfrac{\EigenVal{\lambda}_1}{\EigenVal{\lambda}_k}-1)\cdot \Vect{a}_1 + \cdots + (\tfrac{\EigenVal{\lambda}_{k-1}}{\EigenVal{\lambda}_k}-1)\cdot \Vect{a}_{k-1}\) | \(=\) | \(\Vect{0}\) |
In this last equation, each vector \((\tfrac{\EigenVal{\lambda}_i}{\EigenVal{\lambda}_k}-1)\cdot \Vect{a}_i\) belongs to \(\EigenSpc{\EigenVal{\lambda}_i}\). By induction hypothesis,
| \((\tfrac{\EigenVal{\lambda}_i}{\EigenVal{\lambda}_k}-1)\cdot \Vect{a}_i\) | \(=\) | \(\Vect{0}\) |
We also know that \(\EigenVal{\lambda}_i\neq \EigenVal{\lambda}_k\), for \(1\leq i\leq k-1\). Therefore
| \(\tfrac{\EigenVal{\lambda}_i}{\EigenVal{\lambda}_k} -1\) | \(\neq\) | \(0\) |
for \(1\leq i\leq k-1\). But then \(\Vect{a}_i=\Vect{0}\) for \(1\leq i\leq k-1\). The original equation now yields \(\Vect{a}_k=\Vect{0}\), and this implies the claim.