Exercise 4, Section 6.6 of Hoffman’s Linear Algebra

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Definition: (1) Let $W_1,…,W_k$ be subspaces of the vector space $V$. We say that $W_1,…,W_k$ are independent if $\alpha_1+…+\alpha_n=0$, where $\alpha_i\in W_i$, then each $\alpha_i$ is $0$.

(2) If $V$ is a vector space, a projection of $V$ is a linear operator $E$ on $V$ such that $E^2 = E$.


If $E_1$ and $E_2$ are projections onto independent subspaces, then $E_1 + E_2$ is a projection. True or false?

This problem is vague for me. I don’t know exactly what I have to do. I don’t understand “$E_1$ and $E_2$ are projections onto independent subspaces” sentence. Can you please rephrase this problem?

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The statement is FALSE.

According to the book, a projection on $R$ along $N$ means that $R$ is the range of the projection and $N$ is the null space.

Now, define $E_1$ as below:

$$E_1(1,-1)=(1, -1) , \ E_1(1,2)=(0,0);$$

so $E_1$ is a projection on the subspace spanned by $(1,-1)$, and moreover, $E_1(1,0)=(\frac {2}{3}, \frac{-2}{3})$, $E_1(0,1)=(\frac{-1}{3},\frac{1}{3}).$

Similarly, let's define:

$$E_2(1,0)=(1, 0) , \ E_2(0,1)=(0,0);$$ so $E_2$ is a projection on the subspace spanned by $(1,0).$

Observe that the subspaces spanned by $(1,-1)$ and $(1,0)$ are independent, regarding the definition of the book.

Now, you can easily check that:

$$(E_1+E_2)(1,-1)\neq (E_1+E_2)^2(1,-1).$$


PS: I took this example from this online solution book, Exercise 6.6.4.