Does this notation produce a scalar?

Question

In Blum, Hopcroft and Kannan's "Foundations of Data Science" they discuss the Perceptron Algorithm (p. 111). In the proof it says:

We will keep track of two quantities, w^Tw^* and |w|². Each update increases w^Tw^* by at least 1.

(w + x_il_i)^Tw^* = w^Tw^* + x_i^Tl_iw^* ≥ w^Tw^* + 1

The implication seems that both w^Tw^* and |w|² are scalar values. I interpret |w|² as the dot product of w with itself, as in this answer. However w^Tw^* (and similar uses of transpose throughout the proof) does not make sense to me, as I don't see how it can produce a scalar value.

My reasoning is as follows:

The vector of weights w is a row vector.
The vector of weights w^*, which assumes the "if" condition of the theorem, is also a row vector.

A transposed row vector, multiplied by a row vector, is a column vector multiplied by a row vector, which produces a matrix the size of the column multiplied by the length of the row.

For example, let:

$$\mathbf{w} = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$$ $$\mathbf{w}^* = \begin{bmatrix} 2 & 4 & 6 \end{bmatrix}$$

Then:

$$\mathbf{w}^T\mathbf{w}^* = \begin{bmatrix} 1\\2\\3 \end{bmatrix} \begin{bmatrix} 2 & 4 & 6 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 6\\4 & 8 & 12\\6 & 12 & 18 \end{bmatrix}$$

My assumption is that "≥ w^Tw^* + 1" means "equal to or larger than the sum of two scalars, w^Tw^* + 1". Hence w^Tw^* should be a scalar.

What am I missing?

Thanks in advance.