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The scalar multiplication $\lambda A$ multiplies every one of the $n$ rows by $\lambda$. Applying multilinearity of the determinant once to each row gives

The Leibniz formula for determinants isEquivalently, the determinant is the unique alternating function of the rows that is multilinear and takes value one on the identity matrix. Multiplying one row by $\lambda$ therefore gives

Let the columns be $C_1,\dots ,C_n$, and replace $C_k$ by $C_k+\lambda C_l$, where $k\ne l$. By multilinearity of the determinant,The second determinant vanishes because it has two equal columns. Hence

Expand $\det (A+tB)$ by multilinearity of the determinant in its columns. The coefficient of $t^j$ is a sum of determinants obtained by choosing $j$ columns from $B$ and the remaining columns from $A$. Since the columns of $B$ span a space of dimension $k$, every choice of more than $k$ columns from $B$ is linearly dependent. All coefficients of $t^j$ for $j>k$ therefore vanish, so

If $k=n$, then $B$ is invertible and the coefficient of $t^n$ is $\det B\ne 0$. Hence the degree is exactly $n$.

For a zero-polynomial example with $k=1$, takeThen $\mathrm{rank}B=1$, while $\det (A+tB)=0$ for every $t$.