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I am given a basis $C = \{c_1, ...,c_5\}$ to a real vector space $W$ and an endomorphism $E$ for $W$. I am also given some information about $E$ in the form of $E(c_i) = \lambda_{1,i} \cdot c_1 + \lambda_{2,i} \cdot c_2 + ... + \lambda_{5,i} \cdot c_5$. The exact value of each $E(c_i)$ does not matter for this question. Now I need to check $E$ for bijectivity.

First, I calculated the transformation matrix $A$ by using the information given about $E(c_i)$ and the formula

$E(c_k) = \sum\limits_{i=1}^n(a_{ik} \cdot d_i)$

which can be simplyfied to

$E(c_k) = \sum\limits_{i=1}^5(a_{ik} \cdot c_i)$

since $E$ is an endomorphism and $n=5$.

Beginning the check for bijectivity, I first tried to verify $E$'s injectivity by proving that $E$'s null space $N(E) = \{0\}$. To prove it I calculated $A$'s rank $R(A)$.

It turned out that $R(A) = 4$ although there are no zero rows/columns in $A$ and $A \in \mathbb{R}^{5 \times 5}$. My deduction from that is that there is a vector $v$ so that $A \cdot v = 0, v \neq 0 \Leftrightarrow N(E) \neq \{0\}$. Thus $E$ is not injective and thus not bijective.

My conclusion from $R(A) = 4$ to "$E$ is not bijective" seems a bit vague to me. Is my proof correct and did I show everything that is needed for a proof, or did I skip something?

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    By the formula in your first paragraph, $E(c_1)=E(c_2)=\cdots=E(c_5)$. Is that really what you want? – 2017-01-29
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    $E(c_1) = \sum\limits_{i=1}^5(a_{i1} \cdot c_i) = a_{11} \cdot c_1 + a_{21} \cdot c_2 + ... + a_{51} \cdot c_5$. I then can find each $a_{ik}$ by using the information given about $E(c_i) = \lambda_1 \cdot c_1 + ... + \lambda_5 \cdot c_5$. – 2017-01-29
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    That's not the formula in your first paragraph. The formula in your first paragraph has the same coefficients $\lambda_i$ for each of $c_1,\dots,c_5$. – 2017-01-29
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    Ah sorry, that was a bit of a misformulation. – 2017-01-29
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    Looks OK now. Maybe simpler to say that since the rank is 4 the transformation is not surjective and therefore not bijective. – 2017-01-29

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